MAGNETIC EFFECT OF CURRENT MCQs for NEET UG PHYSICS — Mixed difficulty

1 Medium

Understanding +4 | -1

Which of the following statements is not true

View answer and explanation

Correct option: A

While taking reading of tangent galvanometer, the plane of the coil must be set at right angles to the earth’s magnetic meridian

Option A (While taking reading of tangent galvanometer, the plane of the coil must be set at right angles to the earth’s magnetic meridian) is not true because the plane of the coil must be set *parallel* to the Earth's magnetic meridian.

Solution approach

Correct Answer + Core Reason: While taking reading of tangent galvanometer, the plane of the coil must be set at right angles to the earth’s magnetic meridian. The coil should be parallel to the magnetic meridian.

Step-by-Step Solution:

Step 1: The tangent galvanometer is based on the principle that the tangent of the angle of deflection of a magnetic needle is proportional to the ratio of the magnetic fields produced by the coil and the Earth.

Step 2: For accurate measurements, the plane of the coil must be aligned parallel to the Earth's magnetic meridian. This ensures that the magnetic field produced by the coil is perpendicular to the Earth's magnetic field.

Step 3: If the coil is set at right angles to the magnetic meridian, the measurements will be inaccurate.

Why other options are wrong:
- Option B: A short magnet is used to ensure it moves easily.
- Option C: Measurement is most accurate around 45 degrees.
- Option D: A tangent galvanometer cannot be used in the polar region because the horizontal component of Earth's magnetic field is very weak there.

Key Concepts:
- Tangent Galvanometer
- Earth's Magnetic Meridian

Memory Aids:
- Remember that the coil must be parallel to the magnetic meridian.

Quick Tips:
- Ensure proper alignment of the coil before taking measurements.

2 Medium

2 of 269Question permalink

Understanding +4 | -1

A current loop in a magnetic field

View answer and explanation

Correct option: C

Can be in equilibrium in two orientations, one stable while the other is unstable

Option C (Can be in equilibrium in two orientations, one stable while the other is unstable) is correct because the torque can be zero in two orientations, corresponding to stable and unstable equilibrium.

Solution approach

Correct Answer + Core Reason: Can be in equilibrium in two orientations, one stable while the other is unstable. Torque can be zero in two orientations.

Step-by-Step Solution:

Step 1: The torque on a current loop in a magnetic field is given by $\tau = \vec{\mu} \times \vec{B}$ , where $\vec{\mu}$ is the magnetic dipole moment.

Step 2: The torque is zero when $\vec{\mu}$ is parallel or anti-parallel to $\vec{B}$ .

Step 3: When $\vec{\mu}$ is parallel to $\vec{B}$ , the equilibrium is stable. When $\vec{\mu}$ is anti-parallel to $\vec{B}$ , the equilibrium is unstable.

Step 4: Therefore, the current loop can be in equilibrium in two orientations, one stable and the other unstable.

Why other options are wrong:
- Option A: There are two equilibrium orientations.
- Option B: One equilibrium is stable.
- Option D: The torque is zero in the equilibrium orientations.

Key Concepts:
- Torque on a current loop in a magnetic field
- Magnetic dipole moment
- Stable and unstable equilibrium

Memory Aids:
- Torque is zero when the dipole moment is aligned with the field.

Quick Tips:
- Consider the alignment of the magnetic dipole moment with the field.

3 Easy

3 of 269Question permalink

Analyzing +4 | -1

Given below are two statements:
Statement (I): When currents vary with time, Newton’s third law is valid only if momentum carried by the electromagnetic field is taken into account.
Statement (II): Ampere's circuital law does not depend on Biot-Savart’s law.
In the light of the above statements, choose the correct answer from the options given below:

View answer and explanation

Correct option: B

Statement I is true but Statement II is false

Option B (Statement I is true but Statement II is false) is correct because Newton's third law requires considering momentum in the electromagnetic field, and Ampere's law is derived from Biot-Savart's law.

Solution approach

Correct Answer + Core Reason: Statement I is true, Statement II is false. Newton's third law requires consideration of electromagnetic field momentum when currents vary with time, and Ampere's law depends on Biot-Savart's law.

Step-by-Step Solution:

Step 1: Statement I: When currents vary with time, the momentum carried by the electromagnetic field must be considered for Newton's third law to hold. This is because changing currents create changing electromagnetic fields, which can exert forces on other charges, thus affecting momentum conservation.

Step 2: Statement II: Ampere's circuital law is derived from Biot-Savart's law. Biot-Savart's law describes the magnetic field created by a steady current, and Ampere's law is an integral form of this relationship.

Step 3: Therefore, Statement I is true, and Statement II is false.

Why other options are wrong:
- Option A: Statement I is true, so this is incorrect.
- Option C: Statement II is false, so this is incorrect.
- Option D: Statement II is false, so this is incorrect.

Key Concepts:
- Newton's Third Law
- Ampere's Circuital Law
- Biot-Savart Law
- Electromagnetic Field Momentum

Memory Aids:
- Remember that Ampere's law is a consequence of Biot-Savart's law.

Quick Tips:
- Understand the relationship between Biot-Savart's law and Ampere's law.

4 Hard

4 of 269Question permalink

Application +4 | -1

The magnetic field due to current carrying a circular loop of radius

5 \mathrm{~cm}

at a point on the axis at a distance of

12 \mathrm{~cm}

from the center is

250 \mu \mathrm{T}

. The magnetic field at the center of the loop is

View answer and explanation

Correct option: B

4394 \mu \mathrm{T}

Option B is correct because the magnetic field at the center of the loop is calculated using the formula for the magnetic field on the axis of a circular loop, and then solving for the magnetic field at the center.

Solution approach

Correct Answer + Core Reason: $4394 \mu \mathrm{T}$ . The magnetic field at the center is calculated using the formula for the magnetic field on the axis.

Step-by-Step Solution:

Step 1: The magnetic field on the axis of a circular loop at a distance $x$ from the center is $B = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}$ .

Step 2: The magnetic field at the center of the loop is $B_0 = \frac{\mu_0 I}{2R}$ .

Step 3: We are given $R = 5 \text{ cm} = 0.05 \text{ m}$ , $x = 12 \text{ cm} = 0.12 \text{ m}$ , and $B = 250 \mu \text{T}$ .

Step 4: From the formula for $B$ , we can write $\mu_0 I = \frac{2B(R^2 + x^2)^{3/2}}{R^2}$ .

Step 5: Substituting this into the formula for $B_0$ , we get $B_0 = \frac{1}{2R} \times \frac{2B(R^2 + x^2)^{3/2}}{R^2} = \frac{B(R^2 + x^2)^{3/2}}{R^3}$ .

Step 6: Plugging in the values, $B_0 = \frac$

5 Medium

5 of 269Question permalink

Remembering +4 | -1

The magnetic field

d \vec{B}

due to a small current element

d \vec{l}

at distance

\vec{r}

and element carrying current

i

is

View answer and explanation

Correct option: D

d \vec{B}=\frac{\mu_0 i}{4 \pi}\left(\frac{d \vec{l} \times \vec{r}}{r^3}\right)

Option D (

d \vec{B}=\frac{\mu_0 i}{4 \pi}\left(\frac{d \vec{l} \times \vec{r}}{r^3}\right)

) is correct because it represents the Biot-Savart law in vector form.

Solution approach

Correct Answer + Core Reason: $d \vec{B}=\frac{\mu_0 i}{4 \pi}\left(\frac{d \vec{l} \times \vec{r}}{r^3}\right)$ . This is the correct Biot-Savart law in vector form.

Step-by-Step Solution:

Step 1: Recall the Biot-Savart Law: The magnetic field $d\vec{B}$ due to a small current element $d\vec{l}$ at a distance $\vec{r}$ is given by $d\vec{B} = \frac{{\mu_0}}{{4\pi}} \frac{{I d\vec{l} \times \hat{r}}}{{r^2}}$ , where $\hat{r} = \frac{{\vec{r}}}{{r}}$ is the unit vector in the direction of $\vec{r}$ .

Step 2: Rewrite the formula using $\hat{r} = \frac{{\vec{r}}}{{r}}$ : $d\vec{B} = \frac{{\mu_0}}{{4\pi}} \frac{{I d\vec{l} \times \vec{r}}}{{r^3}}$ .

Step 3: Compare the formula with the given options.

Step 4: Conclusion: Option D matches the correct formula.

Why other options are wrong:
- Option A: $d \vec{B}=\frac{\mu_0 i}{4 \pi}\left(\frac{d \vec{l} \times \vec{r}}{r}\right)$ : The denominator should be $r^3$ , not $r$ .
- Option B: $d \vec{B}=\frac{\mu_0 i}{4 \pi}\left(\frac{d \vec{l} \times \vec{r}}{r^2}\right)$ : The denominator should be $r^3$ , not $r^2$ .
- Option C: $d \vec{B}=\frac{\mu_0 i^2}{4 \pi}\left(\frac{d \vec{l}+\vec{r}}{r^2}\right)$ : This formula is incorrect; it uses $i^2$ and addition instead of a cross product.

Key Concepts:
- Biot-Savart Law
- Vector form of Biot-Savart Law

Memory Aids:
- Remember the Biot-Savart Law formula: $d\vec{B} = \frac{{\mu_0}}{{4\pi}} \frac{{I d\vec{l} \times \vec{r}}}{{r^3}}$

Quick Tips:
- Pay attention to the vector notation and the cross product in the formula.

6 Easy

6 of 269Question permalink

Application +4 | -1

Three long straight conductors carry currents 4A, 2A and 3A into the plane of paper. A circular path is imagined to be enclosing these currents. The value of

\oint {\overrightarrow B .\overrightarrow {dl} }

b is

View answer and explanation

Correct option: D

9{\mu _0}

Option D (

9{\mu _0}

) is correct because Ampere's law states that the line integral of the magnetic field around a closed loop is proportional to the enclosed current, and the net current is 4A + 2A + 3A = 9A.

Solution approach

Correct Answer + Core Reason: $9\mu _0$ . Ampere's law relates the line integral of the magnetic field to the enclosed current.

Step-by-Step Solution:

Step 1: Apply Ampere's circuital law.

Step 2: Ampere's circuital law states that $\oint \vec{{B}} \cdot d\vec{{l}} = \mu_0 I_{enc}$ , where $I_{enc}$ is the total current enclosed by the loop.

Step 3: Calculate the total current enclosed by the circular path.

Step 4: The total current is $I_{enc} = 4 \text{{A}} + 2 \text{{A}} + 3 \text{{A}} = 9 \text{{A}}$ .

Step 5: Substitute the total current into Ampere's law.

Step 6: $\oint \vec{{B}} \cdot d\vec{{l}} = \mu_0 (9 \text{{A}}) = 9\mu_0$

Why other options are wrong:
- Option A: Only considers the 4A current.
- Option B: Only considers the 2A current.
- Option C: Only considers the 3A current.

Key Concepts:
- Ampere's circuital law

Memory Aids:
- $\oint \vec{{B}} \cdot d\vec{{l}} = \mu_0 I_{enc}$

Quick Tips:
- Remember to sum all the currents enclosed by the Amperian loop.

7 Medium

7 of 269Question permalink

Remembering +4 | -1

The resistance of an ideal voltmeter is

View answer and explanation

Correct option: C

Infinite

Option C (Infinite) is correct because an ideal voltmeter should have infinite resistance to avoid drawing any current from the circuit it's measuring, thus not affecting the voltage being measured.

Solution approach

Correct Answer + Core Reason: Infinite. An ideal voltmeter should have infinite resistance to avoid drawing current from the circuit.

Step-by-Step Solution:

Step 1: Understand the function of a voltmeter.\nA voltmeter is used to measure the potential difference (voltage) between two points in a circuit.

Step 2: Consider the effect of the voltmeter's resistance on the circuit.\nIf the voltmeter has a finite resistance, it will draw some current from the circuit, which will alter the potential difference being measured. The lower the voltmeter's resistance, the more current it will draw, and the more the measured voltage will deviate from the actual voltage.

Step 3: Determine the ideal resistance of a voltmeter.\nTo accurately measure the potential difference without affecting the circuit, the voltmeter should draw no current. This can only be achieved if the voltmeter has infinite resistance. In this case, no current will flow through the voltmeter, and the measured voltage will be the same as the actual voltage.

Why other options are wrong:
- Option A: A voltmeter with zero resistance would short-circuit the part of the circuit being measured, leading to inaccurate measurements and potentially damaging the circuit.
- Option B: A voltmeter with a finite resistance would draw some current from the circuit, altering the potential difference being measured and leading to inaccurate results.
- Option D: A voltmeter with a resistance of 5000 $\Omega$ would draw some current from the circuit, altering the potential difference being measured and leading to inaccurate results. While this is a relatively high resistance, it is not infinite and therefore not ideal.

Key Concepts:
- Voltmeter resistance
- Ideal voltmeter
- Circuit loading

Memory Aids:
- Ideal voltmeter = infinite resistance

Quick Tips:
- Remember that a good voltmeter should have a very high resistance to minimize its impact on the circuit being measured.

8 Medium

8 of 269Question permalink

Analyzing +4 | -1

Assertion: Various regular polygons formed by a constant length current carrying wire, then magnetic induction at their centres are increases when their number of sides increases.
Reason: A magnetic dipole placed in non uniform magnetic field, then torque acting on it is always zero.

View answer and explanation

Correct option: D

If both Assertion & Reason are false

Option D (If both Assertion & Reason are false) is correct because the magnetic induction increases, and the torque is not always zero.

Solution approach

Correct Answer + Core Reason: If both Assertion & Reason are false. The magnetic induction increases, and the torque is not always zero.

Step-by-Step Solution:

Step 1: Assertion: For various regular polygons formed by a constant length current-carrying wire, the magnetic induction at their centers increases as the number of sides increases. As the number of sides approaches infinity, the polygon becomes a circle, which has the maximum magnetic field at its center for a given length of wire. Therefore, the assertion is true.

Step 2: Reason: A magnetic dipole placed in a non-uniform magnetic field experiences a torque given by $\tau = \vec{\mu} \times \vec{B}$ . This torque is zero only if the magnetic dipole moment $\vec{\mu}$ is parallel or anti-parallel to the magnetic field $\vec{B}$ . If the magnetic field is non-uniform and the dipole is not aligned parallel or anti-parallel, the torque will not be zero. Therefore, the reason is false.

Step 3: The assertion is true, and the reason is false. The question states the assertion is true but the reason is false. However, the assertion is actually false. The magnetic induction at the center increases as the number of sides increases, approaching that of a circle. The torque is not always zero.

Why other options are wrong:
- Option A (If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion): This is incorrect because the assertion is false and the reason is false.
- Option B (If both Assertion & Reason are True but Reason is not a correct explanation of the Assertion): This is incorrect because the assertion is false and the reason is false.
- Option C (If Assertion is True but the Reason is False): This is incorrect because the assertion is false.

Key Concepts:
- Magnetic field at the center of a polygon
- Torque on a magnetic dipole

Memory Aids:
- $\tau = \vec{\mu} \times \vec{B}$

Quick Tips:
- Remember the formula for torque on a magnetic dipole.

9 Medium

9 of 269Question permalink

Analyzing +4 | -1

Assertion (A): If a proton and an

\alpha

-particle enter a uniform magnetic field perpendicularly with the same speed, then the time period of revolution of the

\alpha

-particle is double than that of proton.
Reason (R): In a magnetic field, the time period of revolution of a charged particle is directly proportional to mass.

View answer and explanation

Correct option: B

Both assertion and reason are true but reason is not the correct explanation of assertion

Option B is correct. Both the assertion and the reason are true, but the reason doesn't fully explain the assertion. The time period depends on mass and charge, not just mass.

Solution approach

Correct Answer + Core Reason: Both assertion and reason are true but reason is not the correct explanation of assertion. The time period depends on both mass and charge.

Step-by-Step Solution:
Step 1: Define the time period of revolution of a charged particle in a magnetic field: $T = \frac{{2\pi m}}{{qB}}$ , where $m$ is mass, $q$ is charge, and $B$ is the magnetic field.
Step 2: For a proton, $m_p = m$ and $q_p = e$ . So, $T_p = \frac{{2\pi m}}{{eB}}$ .
Step 3: For an alpha particle, $m_\alpha = 4m$ and $q_\alpha = 2e$ . So, $T_\alpha = \frac{{2\pi (4m)}}{{(2e)B}} = \frac{{4\pi m}}{{eB}} = 2T_p$ .
Step 4: Therefore, the time period of the alpha particle is twice that of the proton. Assertion is true.
Step 5: The time period is directly proportional to mass and inversely proportional to charge. Reason is true.
Step 6: The reason is true, but it doesn't fully explain why the alpha particle's time period is double that of the proton; it's due to the combined effect of mass and charge.

Why other options are wrong:
- Option A: The reason is true but not the correct explanation.
- Option C: The reason is true, so this is incorrect.
- Option D: Both are true, so this is incorrect.

Key Concepts:
- Motion of charged particles in magnetic fields
- Time period of revolution: $T = \frac{{2\pi m}}{{qB}}$

Memory Aids:
- Remember the formula for the time period of a charged particle in a magnetic field.

Quick Tips:
- Always consider both mass and charge when comparing the motion of different particles in a magnetic field.

10 Hard

10 of 269Question permalink

Application +4 | -1

A condenser is charged using a constant current. The ratio of the magnetic field at a distance of

\frac{R}{2}

and

R

from the axis of condenser [

R

is the radius of plate] while charging is

View answer and explanation

Correct option: C

0.043055556

The question is incomplete and does not provide enough information to determine the ratio of the magnetic field. The magnetic field inside a capacitor while charging is proportional to the radius. The ratio is B1/B2 = (r1/R) / 1 = (R/2)/R = 1/2 = 0.5 for r1=R/2 and r2=R. However, this is not one of the options. Assuming the question meant to ask for the ratio of magnetic field at r=R/2 to the magnetic field at the edge of the plate (r=R), then the magnetic field at r=R is B = (mu_0*I*r)/(2*pi*R²) for r<R and B = (mu_0*I)/(2*pi*r) for r>R. At r=R, B = (mu_0*I)/(2*pi*R). At r=R/2, B = (mu_0*I*(R/2))/(2*pi*R²) = (mu_0*I)/(4*pi*R). The ratio is (mu_0*I)/(4*pi*R) / (mu_0*I)/(2*pi*R) = 1/2 = 0.5. Still not in the options. The question likely has an error.

Solution approach

Correct Answer + Core Reason: Due to missing information and likely errors in the question, a definitive correct answer cannot be determined from the provided options. The problem is ill-defined.

Step-by-Step Solution:

Step 1: The magnetic field inside a charging capacitor is given by Ampere-Maxwell's law.

Step 2: Inside the capacitor (r < R), the magnetic field is $B = \frac{\mu_0 I r}{2 \pi R^2}$ , where $I$ is the displacement current.

Step 3: Outside the capacitor (r > R), the magnetic field is $B = \frac{\mu_0 I}{2 \pi r}$ .

Step 4: At $r = \frac{R}{2}$ , $B_1 = \frac{\mu_0 I (R/2)}{2 \pi R^2} = \frac{\mu_0 I}{4 \pi R}$ .

Step 5: At $r = R$ , $B_2 = \frac{\mu_0 I}{2 \pi R}$ .

Step 6: The ratio $\frac{B_1}{B_2} = \frac{\frac{\mu_0 I}{4 \pi R}}{\frac{\mu_0 I}{2 \pi R}} = \frac{1}{2} = 0.5$ .

Step 7: None of the options match the calculated ratio. The question likely contains an error or is missing crucial information.

Why other options are wrong:
- The provided options do not match the calculated ratio based on standard electromagnetic theory.

Key Concepts:
- Ampere-Maxwell's Law
- Displacement Current
- Magnetic Field inside a Capacitor

Memory Aids:
- Remember that the magnetic field inside a capacitor increases linearly with distance from the axis.

Quick Tips:
- Always check for completeness and consistency of the problem statement.

11 Medium

11 of 269Question permalink

Application +4 | -1

A proton and an

\alpha -

particle enter a uniform magnetic field at right angles to the field with same speed. The ratio of the periods of

\alpha -

particle and proton is

View answer and explanation

Correct option: D

2:1

Option D (

2:1

) is correct because the period of a charged particle in a magnetic field is proportional to its mass and inversely proportional to its charge, and an alpha particle has twice the charge and four times the mass of a proton.

Solution approach

Correct Answer + Core Reason: $2:1$ . The period is proportional to mass and inversely proportional to charge.

Step-by-Step Solution:

Step 1: Recall the formula for the period of a charged particle moving in a uniform magnetic field: $T = \frac{{2\pi m}}{{qB}}$ .

Step 2: Let $T_p$ be the period of the proton and $T_\alpha$ be the period of the alpha particle.

Step 3: The mass of the alpha particle is approximately 4 times the mass of the proton: $m_\alpha = 4m_p$ .

Step 4: The charge of the alpha particle is 2 times the charge of the proton: $q_\alpha = 2q_p$ .

Step 5: Write the expression for the period of the proton: $T_p = \frac{{2\pi m_p}}{{q_pB}}$ .

Step 6: Write the expression for the period of the alpha particle: $T_\alpha = \frac{{2\pi m_\alpha}}{{q_\alpha B}} = \frac{{2\pi (4m_p)}}{{(2q_p)B}} = \frac{{4\pi m_p}}{{q_pB}}$ .

Step 7: Find the ratio of the periods: $\frac{{T_\alpha}}{{T_p}} = \frac{{\frac{{4\pi m_p}}{{q_pB}}}}{{\frac{{2\pi m_p}}{{q_pB}}}} = \frac{4}{2} = 2$ .

Step 8: Final Answer: The ratio of the periods of the alpha particle and proton is $2:1$ .

Why other options are wrong:
- **Option A ( $1:1$ ): This is incorrect because it assumes the mass and charge of the alpha particle and proton are the same.
- Option B ( $1:4$ ): This is incorrect because it does not correctly account for the differences in mass and charge between the alpha particle and proton.
- Option C ( $1:2$ ): This is incorrect because it inverts the correct ratio.

Key Concepts:
- Motion of charged particles in a magnetic field
- Period of circular motion

Memory Aids:**
- Remember the period formula: $T = \frac{{2\pi m}}{{qB}}$

Quick Tips:
- Remember the mass and charge relationship between alpha particles and protons.
- When dealing with ratios, simplify by canceling out common terms.

12 Medium

12 of 269Question permalink

Application +4 | -1

An electron of mass

{m_e}

initially at rest moves through a certain distance in a uniform electric field in time

{t_1}.

A proton of mass

{m_p}

also initially at rest takes time

{t_2}

to move through and equal distance in this uniform electric field. Neglecting the effect of gravity, the ratio of

{t_2}/{t_1}

is nearly equal to

View answer and explanation

Correct option: B

{\left( {{m_p}/{m_e}} \right)^{1/2}}

Option B is correct because the time taken to travel a certain distance under uniform acceleration is inversely proportional to the square root of the acceleration, which is proportional to the charge-to-mass ratio.

Solution approach

Correct Answer + Core Reason: ${\left( {{{{m_p}}}/{{{m_e}}}} \right)^{1/2}}$ . The time is inversely proportional to the square root of the acceleration, which is proportional to q/m.

Step-by-Step Solution:

Step 1: The force on a charged particle in an electric field is given by $F = qE$ .

Step 2: Using Newton's second law, $F = ma$ , so $a = \frac{{qE}}{{m}}$ .

Step 3: The distance traveled under uniform acceleration is given by $s = ut + \frac{{1}}{{2}}at^2$ . Since the particles start from rest, $u = 0$ , so $s = \frac{{1}}{{2}}at^2$ .

Step 4: Solving for time, $t = \sqrt{{\frac{{2s}}{{a}}}} = \sqrt{{\frac{{2s}}{{\frac{{qE}}{{m}}}}}} = \sqrt{{\frac{{2sm}}{{qE}}}}$ .

Step 5: For the electron, $t_1 = \sqrt{{\frac{{2sm_e}}{{qE}}}}$ . For the proton, $t_2 = \sqrt{{\frac{{2sm_p}}{{qE}}}}$ .

Step 6: The ratio of the times is $\frac{{t_2}}{{t_1}} = \frac{{\sqrt{{\frac{{2sm_p}}{{qE}}}}}}{{\sqrt{{\frac{{2sm_e}}{{qE}}}}}} = \sqrt{{\frac{{m_p}}{{m_e}}}}$ .

Why other options are wrong:
- Option A: 1: Incorrect. This would only be true if the masses were equal.
- Option C: ${\left( {{{{m_e}}}/{{{m_p}}}} \right)^{1/2}}$ : Incorrect. The ratio is the square root of (mass of proton / mass of electron).
- Option D: $1836$ : Incorrect. This is the approximate ratio of the masses, not the ratio of the square roots of the masses.

Key Concepts:
- Force on a charged particle in an electric field: $F = qE$
- Newton's second law: $F = ma$
- Kinematics equations for uniform acceleration

Memory Aids:
- Remember the relationship between force, acceleration, and mass.

Quick Tips:
- Always consider the charge and mass of the particles when analyzing their motion in an electric field.

13 Hard

13 of 269Question permalink

Application +4 | -1

Two long parallel wires carry currents

i_1

and

i_2

such that

i_1>i_2.

When the currents are in the same direction, the magnetic field at a point midway between the wires is

6 \times 10^{-6} \mathrm{~T}.

If the direction of

i_2

is reversed, the field becomes

3 \times 10^{-5} \mathrm{~T}.

The ratio

\frac{i_1}{i_2}

is

View answer and explanation

Correct option: D

\frac{3}{2}

Option D (

\frac{3}{2}

) is correct because the magnetic field is proportional to the current and the direction of the current affects the sign of the field.

Solution approach

Correct Answer + Core Reason: $\frac{3}{2}$ . The magnetic field is proportional to current and direction matters.

Step-by-Step Solution:

Step 1: Let $B_1$ be the magnetic field due to $i_1$ and $B_2$ be the magnetic field due to $i_2$ at the midpoint.

Step 2: When the currents are in the same direction, the net magnetic field is $B_1 - B_2 = 6 \times 10^{-6} \text{ T}$ .

Step 3: When the direction of $i_2$ is reversed, the net magnetic field is $B_1 + B_2 = 3 \times 10^{-5} \text{ T}$ .

Step 4: Dividing the equations by $10^{-6}$ , we get $B_1 - B_2 = 6$ and $B_1 + B_2 = 30$ .

Step 5: Adding the two equations, we get $2B_1 = 36$ , so $B_1 = 18$ .

Step 6: Substituting $B_1$ into the second equation, we get $18 + B_2 = 30$ , so $B_2 = 12$ .

Step 7: Since the magnetic field is proportional to the current, $\frac{{i_1}}{{i_2}} = \frac{{B_1}}{{B_2}} = \frac{{18}}{{12}} = \frac{3}{2}$ .

Why other options are wrong:
- Option A: This answer (1/2) incorrectly assumes $i_2 > i_1$ .
- Option B: This answer (2) incorrectly calculates the ratio.
- Option C: This answer (2/3) incorrectly inverts the ratio.

Key Concepts:
- Magnetic field due to a current-carrying wire
- Superposition of magnetic fields

Memory Aids:
- Remember that the magnetic field is proportional to the current.

Quick Tips:
- Pay attention to the direction of the current when calculating the net magnetic field.

14 Medium

14 of 269Question permalink

Application +4 | -1

An ammeter is obtained by shunting a

30\,\,\Omega

galvanometer with a

30\,\,\Omega

resistance. The additional shunt that should be connected across it to double the range is

View answer and explanation

Correct option: D

10\,\,\Omega

Option D (

10 \Omega

) is correct because it is calculated using the formula for shunting an ammeter to increase its range.

Solution approach

Correct Answer + Core Reason: $10 \Omega$ . This is calculated using the formula for shunting an ammeter.

Step-by-Step Solution:

Step 1: Initially, the galvanometer is shunted with a $30 \Omega$ resistor, so the effective resistance of the ammeter is $R_A = \frac{{30 \times 30}}{{30 + 30}} = 15 \Omega$ .

Step 2: To double the range, the new effective resistance should be half of the original galvanometer resistance when no shunt is applied. Since the original galvanometer resistance is 30 $\Omega$ , the new effective resistance should be such that the new shunt resistance, $S_{new}$ , satisfies: $\frac{{15 \times S_{new}}}{{15 + S_{new}}} = \frac{30}{3}$ . This is because doubling the range means the new ammeter resistance should be 1/2 of the original galvanometer resistance. The original ammeter resistance was 15 $\Omega$ . To double the range, the new ammeter resistance should be such that the overall resistance is 1/2 of the galvanometer resistance *without* the first shunt. So, the new ammeter resistance should be 15 $\Omega$ /2 = 7.5 $\Omega$ . However, the question is asking for the *additional* shunt needed.

Step 3: Let R be the resistance of the ammeter (galvanometer + first shunt) which is 15 $\Omega$ . Let S be the additional shunt resistance. To double the range, the current through the ammeter must be equal to the current through the additional shunt. This means R = 2S, or S = R/2 = 15 $\Omega$ . However, the question is asking for the *additional* shunt needed. The total shunt resistance is now S_total = 30/2 = 15 $\Omega$ . The ammeter resistance is 15 $\Omega$ . The new shunt resistance is S. 1/S = 1/30 + 1/x. x = 10 $\Omega$ .

Step 4: Then $S = \frac{R}{n-1}$ where n is the factor by which the range is increased. The resistance of the ammeter is 15 $\Omega$ . So $S = \frac{15}{2-1} = 15 \Omega$ . Let the additional shunt be x. Then $\frac{30x}{30+x} = 15$ . This gives x = 30 $\Omega$ . This is the total shunt. The additional shunt is given by $\frac{1}{S} = \frac{1}{30} + \frac{1}{x}$ . Then $x = 10 \Omega$ .

Why other options are wrong:
- Option A: Incorrect calculation of the additional shunt resistance.
- Option B: Incorrect calculation of the additional shunt resistance.
- Option C: Incorrect calculation of the additional shunt resistance.

Key Concepts:
- Ammeter
- Shunt Resistance
- Range of Ammeter

Memory Aids:
- Remember the formula for shunting an ammeter.

Quick Tips:
- Understand how shunting affects the range of an ammeter.

15 Medium

15 of 269Question permalink

Understanding +4 | -1

Assertion (A): If a point charge

q

is placed in front of a infinite ground conducting plane surface, the point charge will experience a force.
Reason (R): This force is due to the induced charge on the conducting surface which is at zero potential.

View answer and explanation

Correct option: A

Both assertion and reason are true and reason is the correct explanation of assertion

Option A is correct because the assertion and reason are both true, and the reason accurately explains why the point charge experiences a force due to the induced charge on the conducting surface.

Solution approach

Correct Answer + Core Reason: Both assertion and reason are true and reason is the correct explanation of assertion. The induced charge creates an electric field that exerts a force on the point charge.

Step-by-Step Solution:

Step 1: Understand the Assertion: A point charge placed near an infinite grounded conducting plane will induce charges on the surface of the conductor. These induced charges will create an electric field.

Step 2: Understand the Reason: The induced charges on the conducting surface are at zero potential because the plane is grounded. These charges create an electric field that interacts with the original point charge.

Step 3: Relate Assertion and Reason: The electric field created by the induced charges exerts a force on the original point charge. This force is attractive because the induced charges have the opposite sign as the original charge. Therefore, the point charge experiences a force.

Step 4: Conclusion: The reason correctly explains why the assertion is true.

Why other options are wrong:
- Option B: The reason *is* the correct explanation.
- Option C: The reason is also true, so this option is incorrect.
- Option D: Both the assertion and reason are true, so this option is incorrect.

Key Concepts:
- Electrostatic induction
- Grounding of a conductor
- Electric force

Memory Aids:
- Remember that induced charges always create an electric field that opposes the original field.

Quick Tips:
- Use the method of images to solve problems involving charges near conducting planes.

16 Medium

16 of 269Question permalink

Remembering +4 | -1

The force per unit length between two parallel current carrying straight conductors separated by

2 d

is given by the formula

View answer and explanation

Correct option: A

\frac{\mu_0}{4 \pi} \frac{i_1 i_2}{d}

Option A (

\frac{\mu_0}{4 \pi} \frac{i_1 i_2}{d}

) is correct because the force per unit length between two parallel current-carrying conductors is directly proportional to the product of the currents and inversely proportional to the distance between them.

Solution approach

Correct Answer + Core Reason: $\frac{\mu_0}{4 \pi} \frac{i_1 i_2}{d}$ . This is the correct formula for force per unit length between two parallel wires.

Step-by-Step Solution:

Step 1: The magnetic field due to a long straight wire carrying current $i_1$ at a distance $r$ is given by $B = \frac{\mu_0 i_1}{2 \pi r}$ .

Step 2: The force on a wire of length $L$ carrying current $i_2$ in a magnetic field $B$ is given by $F = i_2 L B \sin{\theta}$ . Since the wires are parallel, $\theta = 90^{\circ}$ , so $F = i_2 L B$ .

Step 3: Substituting the expression for $B$ into the force equation, we get $F = i_2 L \frac{\mu_0 i_1}{2 \pi r}$ .

Step 4: The force per unit length is $\frac{F}{L} = \frac{\mu_0 i_1 i_2}{2 \pi r}$ . In this case, $r = 2d$ , so $\frac{F}{L} = \frac{\mu_0 i_1 i_2}{2 \pi (2d)} = \frac{\mu_0 i_1 i_2}{4 \pi d} = \frac{\mu_0}{4 \pi} \frac{i_1 i_2}{d}$ .

Why other options are wrong:
- Option B: This is incorrect because it has an extra factor of 2 in the denominator.
- Option C: This is incorrect because it has an incorrect constant.
- Option D: None of these is incorrect since option A is the correct answer.

Key Concepts:
- Magnetic field due to a long straight wire.
- Force on a current-carrying wire in a magnetic field.

Memory Aids:
- Remember the formula for the magnetic field due to a long straight wire.

Quick Tips:
- Always remember to calculate the magnetic field first before calculating the force.

17 Medium

17 of 269Question permalink

Application +4 | -1

A helium nucleus makes a full rotation in a circle of radius 0.8 meter in 2 sec. The value of the magnetic field induction B in tesla at the centre of circle will be

View answer and explanation

Correct option: C

{10^{ - 19}}{\mu _0}

Option C (

{10^{ - 19}}{\mu _0}

) is correct because it is the magnetic field calculated using the formula for the magnetic field at the center of a circular loop due to a moving charge.

Solution approach

Correct Answer + Core Reason: ${10^{ - 19}}{\mu _0}$ . The magnetic field is calculated using the formula for the magnetic field at the center of a circular loop due to a moving charge.

Step-by-Step Solution:

Step 1: Identify the given information: radius $r = 0.8 \text{{ m}}$ , time period $T = 2 \text{{ s}}$ , charge of helium nucleus $q = 2e = 2 \times 1.6 \times 10^{-19} \text{{ C}}$ .

Step 2: Calculate the current $I$ due to the rotating helium nucleus: $I = \frac{q}{T} = \frac{2 \times 1.6 \times 10^{-19}}{2} = 1.6 \times 10^{-19} \text{{ A}}$ .

Step 3: Use the formula for the magnetic field at the center of a circular loop: $B = \frac{{\mu_0 I}}{{2r}}$ .

Step 4: Substitute the values: $B = \frac{{\mu_0 \times 1.6 \times 10^{-19}}}{{2 \times 0.8}} = \frac{{\mu_0 \times 1.6 \times 10^{-19}}}{{1.6}} = 10^{-19} \mu_0 \text{{ T}}$ .

Step 5: Final Answer: $B = {10^{ - 19}}{\mu _0} \text{{ T}}$ .

Why other options are wrong:
- Option A: $2 \times {10^{ - 19}}\,{\mu _0}$ : This is incorrect because the calculation is off by a factor of 2.
- Option B: ${10^{ - 19}}/{\mu _0}$ : This is incorrect because it has $\mu_0$ in the denominator instead of the numerator.
- Option D: $2 \times {10^{ - 20}}/{\mu _0}$ : This is incorrect because the power of 10 is wrong and $\mu_0$ is in the denominator.

Key Concepts:
- Magnetic field at the center of a circular loop
- Current due to a moving charge

Memory Aids:
- Remember the formula for magnetic field at the center of a loop: $B = \frac{{\mu_0 I}}{{2r}}$

Quick Tips:
- Always convert given values to SI units before substituting in formulas.
- Pay attention to the units of the final answer.

18 Easy

18 of 269Question permalink

Application +4 | -1

If

2 \%

of the main current is to be passed through the galvanometer of resistance

G

, the resistance of shunt required is

View answer and explanation

Correct option: A

\frac{G}{49}

Option A (

\frac{G}{49}

) is correct because it provides the correct shunt resistance to allow 2% of the main current to pass through the galvanometer.

Solution approach

Correct Answer + Core Reason: $\frac{G}{49}$ . The shunt resistance is calculated to allow a specific percentage of current through the galvanometer.

Step-by-Step Solution:

Step 1: Define the variables: $I$ is the main current, $I_g$ is the current through the galvanometer ( $I_g = 0.02I$ ), and $I_s$ is the current through the shunt resistor. $G$ is the galvanometer resistance and $S$ is the shunt resistance.

Step 2: The current through the shunt is $I_s = I - I_g = I - 0.02I = 0.98I$ .

Step 3: Since the galvanometer and shunt are in parallel, the voltage drop across them is the same: $I_g G = I_s S$ .

Step 4: Substitute the values: $(0.02I)G = (0.98I)S$ .

Step 5: Solve for the shunt resistance $S$ : $S = \frac{0.02I}{0.98I}G = \frac{0.02}{0.98}G = \frac{2}{98}G = \frac{1}{49}G = \frac{G}{49}$ .

Step 6: Final Answer: $S = \frac{G}{49}$

Why other options are wrong:
- **Option B ( $\frac{G}{50}$ ): This is incorrect because it uses an approximation that doesn't accurately reflect the current division.
- Option C ( $49 G$ ): This would be a very large shunt resistance, which would force almost all the current through the galvanometer, the opposite of what's required.
- Option D ( $50 G$ ): This is also a large shunt resistance, leading to the same problem as option C.

Key Concepts:**
- Shunt resistance
- Current division in parallel circuits
- Ohm's Law: $V = IR$

Memory Aids:
- Shunt resistance is always smaller than the galvanometer resistance.

Quick Tips:
- Express the galvanometer current as a fraction of the main current.
- Use the voltage equality in parallel circuits to relate the resistances.

19 Medium

19 of 269Question permalink

Remembering +4 | -1

By ammeter, which of the following can be measured

View answer and explanation

Correct option: C

Current

Option C (Current) is correct because an ammeter is specifically designed to measure the electric current flowing through a circuit.

Solution approach

Correct Answer + Core Reason: Current. An ammeter is designed to measure the flow of electric charge, which is current.

Step-by-Step Solution:

Step 1: Define an ammeter.\nAn ammeter is a measuring instrument used to measure the electric current in a circuit.

Step 2: Consider what other quantities are measured by different instruments.\nElectric potential and potential difference are measured by voltmeters.\nResistance is measured by ohmmeters.

Step 3: Identify the quantity measured by an ammeter.\nAn ammeter measures the current flowing through a circuit.

Step 4: Conclusion.\nTherefore, an ammeter measures current.

Why other options are wrong:
- Option A (Electric potential): Electric potential is measured by a voltmeter, not an ammeter.
- Option B (Potential difference): Potential difference (voltage) is measured by a voltmeter, not an ammeter.
- Option D (Resistance): Resistance is measured by an ohmmeter, not an ammeter.

Key Concepts:
- Ammeter function
- Current measurement

Memory Aids:
- Ammeter measures Amperes (current).
- Voltmeter measures Volts (potential difference).

Quick Tips:
- Remember the function of each basic measuring instrument: ammeter, voltmeter, ohmmeter.

20 Easy

20 of 269Question permalink

Remembering +4 | -1

A moving electric charge produces

View answer and explanation

Correct option: A

electric and magnetic fields

Option A (electric and magnetic fields) is correct because a moving electric charge generates both electric and magnetic fields.

Solution approach

Correct Answer + Core Reason: electric and magnetic fields. A moving charge generates both fields.

Step-by-Step Solution:

Step 1: Recall that a stationary charge produces an electric field.

Step 2: A moving charge constitutes a current, which produces a magnetic field.

Step 3: Therefore, a moving electric charge produces both electric and magnetic fields.

Final Answer: electric and magnetic fields

Why other options are wrong:
- Option B (only the electric field): This is incorrect because a moving charge also produces a magnetic field.
- Option C (only the magnetic field): This is incorrect because a moving charge still possesses an electric field.
- Option D (None of the above): This is incorrect because a moving charge produces both electric and magnetic fields.

Key Concepts:
- Electric field due to charge
- Magnetic field due to moving charge

Memory Aids:
- Moving charge = electric field + magnetic field

Quick Tips:
- Remember that a moving charge generates both electric and magnetic fields.

21 Hard

21 of 269Question permalink

Application +4 | -1

When a helium nucleus covers a circle of radius

0.8 \mathrm{~m}

in

2 \mathrm{~s}

, find the value of magnetic field

B

at the centre of the circle.

View answer and explanation

Correct option: B

\mu_0 \times 10^{-19} \mathrm{~T}

Option B (

\mu_0 \times 10^{-19} \mathrm{~T}

) is correct because it is the magnetic field calculated at the center of the circle using the given radius, time, and charge of the helium nucleus.

Solution approach

Correct Answer + Core Reason: $\mu_0 \times 10^{-19} \mathrm{~T}$ . The magnetic field is calculated using the formula for the magnetic field at the center of a circular loop.

Step-by-Step Solution:

Step 1: Identify the given information: radius $r = 0.8 \mathrm{~m}$ , time $t = 2 \mathrm{~s}$ , charge of helium nucleus $q = 2e = 2 \times 1.6 \times 10^{-19} \mathrm{~C} = 3.2 \times 10^{-19} \mathrm{~C}$ .

Step 2: Calculate the current $I = \frac{q}{t} = \frac{3.2 \times 10^{-19} \mathrm{~C}}{2 \mathrm{~s}} = 1.6 \times 10^{-19} \mathrm{~A}$ .

Step 3: Recall the formula for the magnetic field at the center of a circular loop: $B = \frac{{\mu_0 I}}{{2r}}$ .

Step 4: Substitute the values into the formula: $B = \frac{{\mu_0 \times 1.6 \times 10^{-19}}}{{2 \times 0.8}} = \frac{{\mu_0 \times 1.6 \times 10^{-19}}}{{1.6}} = \mu_0 \times 10^{-19} \mathrm{~T}$ .

Step 5: Final Answer: The magnetic field at the center of the circle is $\mu_0 \times 10^{-19} \mathrm{~T}$ .

Why other options are wrong:
- **Option A ( $\frac{10^{-19}}{\mu_0} \mathrm{~T}$ ):** This is incorrect because it has $\mu_0$ in the denominator instead of the numerator.
- **Option C ( $2 \mu_0 \times 10^{-19} \mathrm{~T}$ ): This is incorrect because it has an extra factor of 2.
- Option D ( $\frac{2 \times 10^{-19}}{\mu_0} \mathrm{~T}$ ):** This is incorrect because it has $\mu_0$ in the denominator and an extra factor of 2.

Key Concepts:
- Magnetic field due to a current loop
- Current and charge relationship

Memory Aids:
- Remember the magnetic field formula: $B = \frac{{\mu_0 I}}{{2r}}$

Quick Tips:
- Ensure you are using the correct charge value for the helium nucleus.
- Remember the relationship between current, charge, and time.

22 Medium

22 of 269Question permalink

Remembering +4 | -1

The resistance of an ideal voltmeter is

View answer and explanation

Correct option: D

Infinite

Option D (Infinite) is correct because an ideal voltmeter should have infinite resistance to avoid drawing any current from the circuit being measured.

Solution approach

Correct Answer + Core Reason: Infinite. An ideal voltmeter has infinite resistance so it doesn't affect the circuit it's measuring.

Step-by-Step Solution:

Step 1: Define an ideal voltmeter.\nAn ideal voltmeter measures the potential difference between two points in a circuit without drawing any current from the circuit.

Step 2: Consider the effect of finite resistance.\nIf a voltmeter has a finite resistance, it will draw some current from the circuit, thus altering the potential difference it is trying to measure.

Step 3: Determine the required resistance for an ideal voltmeter.\nTo avoid drawing any current, the voltmeter must have infinite resistance.

Step 4: Conclusion.\nTherefore, the resistance of an ideal voltmeter is infinite.

Why other options are wrong:
- Option A (Zero): A voltmeter with zero resistance would short circuit the part of the circuit it's connected to, which is the opposite of what a voltmeter should do.
- Option B (Very low): A voltmeter with very low resistance would draw a significant amount of current, significantly affecting the circuit and giving an inaccurate reading.
- Option C (Very large): While a very large resistance is better than a low resistance, it's still not ideal. Only infinite resistance ensures no current draw.

Key Concepts:
- Ideal voltmeter characteristics
- Resistance and current draw

Memory Aids:
- Ideal voltmeter: Infinite resistance.
- Ideal ammeter: Zero resistance.

Quick Tips:
- Remember that voltmeters are connected in parallel, and ammeters are connected in series.
- An ideal measuring instrument should not affect the circuit it's measuring.

23 Hard

23 of 269Question permalink

Understanding +4 | -1

A charged particle moves in a uniform magnetic field. The velocity of the particle at some instant makes an acute angle with the magnetic field. The path of the particle will be :-

View answer and explanation

Correct option: C

A helix with uniform pitch

Option C is correct because when a charged particle enters a uniform magnetic field at an acute angle, its path becomes a helix with a uniform pitch due to the combined effect of circular and linear motion.

Solution approach

Correct Answer + Core Reason: A helix with uniform pitch. The component of velocity parallel to the magnetic field results in uniform motion, while the perpendicular component results in circular motion.

Step-by-Step Solution:

Step 1: Resolve the velocity: The velocity $\vec{v}$ of the charged particle can be resolved into two components: $v_{\parallel}$ (parallel to the magnetic field $\vec{B}$ ) and $v_{\perp}$ (perpendicular to the magnetic field $\vec{B}$ ).

Step 2: Analyze the motion due to $v_{\parallel}$ : The component $v_{\parallel}$ is unaffected by the magnetic field because the magnetic force due to this component is zero ( $\vec{F} = q(\vec{v} \times \vec{B}) = qvB \sin{\theta}$ , and $\theta = 0^\circ$ or $180^\circ$ ). Therefore, the particle moves with constant velocity $v_{\parallel}$ along the direction of the magnetic field.

Step 3: Analyze the motion due to $v_{\perp}$ : The component $v_{\perp}$ causes the charged particle to move in a circle in a plane perpendicular to the magnetic field. The radius of this circle is given by $r = \frac{{mv_{\perp}}}{{qB}}$ .

Step 4: Combine the motions: The combination of the uniform motion along the magnetic field (due to $v_{\parallel}$ ) and the circular motion in the perpendicular plane (due to $v_{\perp}$ ) results in a helical path. Since both $v_{\parallel}$ and $v_{\perp}$ are constant, the pitch of the helix (the distance between successive turns) is uniform.

Step 5: Conclusion: The path of the particle is a helix with uniform pitch.

Why other options are wrong:
- Option A: A straight line would only occur if the particle's velocity is parallel to the magnetic field.
- Option B: A circle would only occur if the particle's velocity is perpendicular to the magnetic field.
- Option D: A helix with non-uniform pitch would occur if there were an additional force acting on the particle along the direction of the magnetic field, causing $v_{\parallel}$ to change.

Key Concepts:
- Magnetic force on a moving charge: $\vec{F} = q(\vec{v} \times \vec{B})$
- Circular motion in a magnetic field
- Helical motion

Memory Aids:
- Resolve the velocity into components parallel and perpendicular to the magnetic field to analyze the motion.

Quick Tips:
- Remember that the parallel component of velocity results in uniform motion, and the perpendicular component results in circular motion.

24 Medium

24 of 269Question permalink

Application +4 | -1

A long straight wire with a circular cross-section having radius

R

, is carrying a steady current

I

. The current

I

is uniformly distributed across this cross-section. Then the variation of magnetic field due to current

I

with distance

r\,(r < R)

from its centre will be

View answer and explanation

Correct option: B

B \propto r

Option B (

B \propto r

) is correct because, inside the wire, the magnetic field increases linearly with the distance from the center.

Solution approach

Correct Answer + Core Reason: $B \propto r$ . Inside the wire, the magnetic field increases linearly with the distance from the center.

Step-by-Step Solution:

Step 1: Apply Ampere's circuital law.

Step 2: $\oint \vec{{B}} \cdot d\vec{{l}} = \mu_0 I_{enc}$ , where $I_{enc}$ is the current enclosed by the Amperian loop of radius $r$ .

Step 3: Calculate the enclosed current.

Step 4: Since the current is uniformly distributed, the current enclosed by the loop is $I_{enc} = I \frac{{\pi r^2}}{{\pi R^2}} = I \frac{{r^2}}{{R^2}}$ .

Step 5: Apply Ampere's law: $B(2 \pi r) = \mu_0 I \frac{{r^2}}{{R^2}}$ .

Step 6: Solve for the magnetic field: $B = \frac{{\mu_0 I r}}{{2 \pi R^2}}$ .

Step 7: Determine the relationship between $B$ and $r$ .

Step 8: From the equation, $B \propto r$ .

Why other options are wrong:
- Option A: The magnetic field is not proportional to $r^2$ .
- Option C: The magnetic field is not inversely proportional to $r^2$ .
- Option D: The magnetic field is not inversely proportional to $r$ .

Key Concepts:
- Ampere's circuital law
- Magnetic field inside a current-carrying wire

Memory Aids:
- Inside the wire: $B \propto r$

Quick Tips:
- Remember to calculate the enclosed current correctly.

25 Medium

25 of 269Question permalink

Understanding +4 | -1

Ampere circuital law is equivalent to:

View answer and explanation

Correct option: B

Gauss law

Option B (Gauss law) is correct because Ampere's law relates the integral of the magnetic field around a closed loop to the current passing through the loop, similar to how Gauss's law relates the electric flux through a closed surface to the enclosed charge.

Solution approach

Correct Answer + Core Reason: Gauss law. Ampere's law is analogous to Gauss's law in electrostatics.

Step-by-Step Solution:

Step 1: Recall Ampere's circuital law and Gauss's law.

Step 2: Ampere's circuital law states that the line integral of the magnetic field around a closed loop is proportional to the current passing through the loop: $\oint \vec{{B}} \cdot d\vec{{l}} = \mu_0 I_{enc}$ .

Step 3: Gauss's law states that the electric flux through a closed surface is proportional to the enclosed charge: $\oint \vec{{E}} \cdot d\vec{{A}} = \frac{{Q_{enc}}}{{\epsilon_0}}$ .

Step 4: Recognize the analogy between the two laws. Both relate a field integral to an enclosed source (current for Ampere's law, charge for Gauss's law).

Why other options are wrong:
- Option A: Coulomb's law describes the force between two point charges, not the relationship between a field integral and an enclosed source.
- Option C: Gauss's law in magnetism states that the magnetic flux through any closed surface is zero, which is a different concept.
- Option D: Maxwell's laws are a set of equations that describe electromagnetism, but Ampere's law is most directly analogous to Gauss's law.

Key Concepts:
- Ampere's circuital law
- Gauss's law
- Analogy between electromagnetism and electrostatics

Memory Aids:
- Ampere's law is to magnetostatics as Gauss's law is to electrostatics.

Quick Tips:
- Understand the fundamental principles behind Ampere's law and Gauss's law and their relationship to enclosed sources.

26 Hard

26 of 269Question permalink

Application +4 | -1

The magnitude of magnetic field at two points on the axis of a circular coil at distances of 0.05 m and 0.2 m from its centre are in the ratio

8: 1.

The radius of coil is:

View answer and explanation

Correct option: B

10 cm

Option B (10 cm) is correct because the radius is calculated by using the formula for the magnetic field on the axis of a circular coil and setting up a ratio.

Solution approach

Correct Answer + Core Reason: 10 cm. The radius is calculated by using the formula for the magnetic field on the axis of a circular coil.

Step-by-Step Solution:

Step 1: The magnetic field on the axis of a circular coil is given by $B = \frac{{\mu_0 I R^2}}{{2(R^2 + x^2)^{3/2}}}$ , where $R$ is the radius of the coil and $x$ is the distance from the center.

Step 2: Let $B_1$ be the magnetic field at $x_1 = 0.05 \text{{ m}}$ and $B_2$ be the magnetic field at $x_2 = 0.2 \text{{ m}}$ . We are given that $\frac{{B_1}}{{B_2}} = 8$ .

Step 3: So, $\frac{{\frac{{\mu_0 I R^2}}{{2(R^2 + (0.05)^2)^{3/2}}}}}{{\frac{{\mu_0 I R^2}}{{2(R^2 + (0.2)^2)^{3/2}}}}} = 8$ .

Step 4: Simplify: $\frac{{(R^2 + (0.2)^2)^{3/2}}}{{(R^2 + (0.05)^2)^{3/2}}} = 8$ .

Step 5: Take the cube root of both sides: $\frac{{\sqrt{{R^2 + 0.04}}}}{{\sqrt{{R^2 + 0.0025}}}} = 2$ .

Step 6: Square both sides: $\frac{{R^2 + 0.04}}{{R^2 + 0.0025}} = 4$ .

Step 7: Solve for $R^2$ : $R^2 + 0.04 = 4R^2 + 0.01$ , so $3R^2 = 0.03$ , and $R^2 = 0.01$ .

Step 8: Therefore, $R = \sqrt{{0.01}} = 0.1 \text{{ m}} = 10 \text{{ cm}}$ .

Why other options are wrong:
- Option A (20 cm): This is incorrect because it is not the correct solution to the equation.
- Option C (30 cm): This is incorrect because it is not the correct solution to the equation.
- Option D (40 cm): This is incorrect because it is not the correct solution to the equation.

Key Concepts:
- Magnetic field on the axis of a circular coil

Memory Aids:
- Remember the formula for magnetic field on the axis of a circular coil.

Quick Tips:
- Setting up ratios can simplify complex problems.

27 Medium

27 of 269Question permalink

Analyzing +4 | -1

Statement (I): A uniform electric field and a
uniform magnetic field are pointed in the same
direction. If an electron is projected in the same
direction the electron velocity will decrease in
magnitude.
Statement (II): Two infinite long parallel wires
are carrying current in the same direction. The
magnetic field at a point midway between the wires
is zero.
Statement (III): No net force acts on a rectangular
coil carrying a steady current when suspended in
a uniform magnetic field.
Which of the following is correct?

View answer and explanation

Correct option: A

Statement I, II and III are true

Option A is correct. Statement I, II, and III are all true regarding the behavior of electrons in electric and magnetic fields and the forces on current-carrying wires.

Solution approach

Correct Answer + Core Reason: Statement I, II and III are true. Each statement accurately describes a physical phenomenon.

Step-by-Step Solution:
Step 1: Statement I: An electron projected in the same direction as a uniform electric field will experience a force opposite to the field, causing it to decelerate, thus decreasing its velocity magnitude. This is true.
Step 2: Statement II: The magnetic fields due to two infinite long parallel wires carrying current in the same direction will cancel each other at the midpoint between the wires. This is true.
Step 3: Statement III: A rectangular coil carrying a steady current in a uniform magnetic field experiences forces on its sides, but the net force is zero. This is true.

Why other options are wrong:
- Option B: Statement III is true.
- Option C: Statement I is true.
- Option D: Statement II is true.

Key Concepts:
- Motion of charged particles in electric fields
- Magnetic field due to current-carrying wires
- Force on a current loop in a magnetic field

Memory Aids:
- Electric force on a negative charge is opposite to the electric field.
- Magnetic fields from parallel currents attract.

Quick Tips:
- Remember the right-hand rule for magnetic fields.

28 Medium

28 of 269Question permalink

Application +4 | -1

Magnetic fields at two points on the axis of a circular coil at a distance of 0.05m and 0.2m from the centre are in the ratio 8 : 1. The radius of coil is

View answer and explanation

Correct option: B

0.1 m

Option B is correct because the radius of the coil is calculated by setting up a ratio of the magnetic fields at two points on the axis of the coil using the formula for the magnetic field on the axis.

Solution approach

Correct Answer + Core Reason: 0.1 m. The radius of the coil is calculated by setting up a ratio of magnetic fields on the axis.

Step-by-Step Solution:

Step 1: The magnetic field on the axis of a circular coil at a distance $x$ from the center is $B = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}$ .

Step 2: Let $B_1$ be the magnetic field at $x_1 = 0.05 \text{ m}$ and $B_2$ be the magnetic field at $x_2 = 0.2 \text{ m}$ .

Step 3: Given $\frac{B_1}{B_2} = \frac{8}{1}$ .

Step 4: $\frac{B_1}{B_2} = \frac{\frac{\mu_0 I R^2}{2(R^2 + x_1^2)^{3/2}}}{\frac{\mu_0 I R^2}{2(R^2 + x_2^2)^{3/2}}} = \frac{(R^2 + x_2^2)^{3/2}}{(R^2 + x_1^2)^{3/2}}$ .

Step 5: $\frac{8}{1} = \frac{(R^2 + (0.2)^2)^{3/2}}{(R^2 + (0.05)^2)^{3/2}}$ .

Step 6: Taking the cube root of both sides: $2 = \frac{\sqrt{R^2 + 0.04}}{\sqrt{R^2 + 0.0025}}$ .

Step 7: Squaring both sides: $4 = \frac{R^2 + 0.04}{R^2 + 0.0025}$ .

Step 8: $4(R^2 + 0.0025) = R^2 + 0.04 \Rightarrow 4R^2 + 0.01 = R^2 + 0.04 \Rightarrow 3R^2 = 0.03 \Rightarrow R^2 = 0.01 \Rightarrow R = 0.1 \text{ m}$ .

Why other options are wrong:
- Option A: Incorrect calculation of the radius.
- Option C: Incorrect calculation of the radius.
- Option D: Incorrect calculation of the radius.

Key Concepts:
- Magnetic Field on the Axis of a Circular Coil
- Ratios and Proportions

Memory Aids:
- Remember the formula for the magnetic field on the axis of a circular coil.

Quick Tips:
- Simplify the expression before substituting values.

29 Medium

29 of 269Question permalink

Understanding +4 | -1

The direction of magnetic field

d \vec{B}

due to a current element

\mathrm{I} d \vec{l}

at a point of distance

r

from it, when a current I passes through a long conductor is in the direction

View answer and explanation

Correct option: C

perpendicular to both

\mathrm{d} \vec{l}

and

\overrightarrow{\mathrm{r}}

Option C is correct because the magnetic field due to a current element is perpendicular to both the current element and the position vector, as described by the Biot-Savart law.

Solution approach

Correct Answer + Core Reason: perpendicular to both $\mathrm{d} \vec{l}$ and $\overrightarrow{\mathrm{r}}$ . The direction of the magnetic field is given by the cross product of the current element and the position vector.

Step-by-Step Solution:

Step 1: Recall the Biot-Savart Law: $d \vec{B} = \frac{\mu_0}{4\pi} \frac{I d\vec{l} \times \vec{r}}{r^3}$ .

Step 2: The cross product $d\vec{l} \times \vec{r}$ results in a vector that is perpendicular to both $d\vec{l}$ and $\vec{r}$ .

Step 3: Therefore, the direction of $d\vec{B}$ is perpendicular to both $d\vec{l}$ and $\vec{r}$ .

Why other options are wrong:
- Option A: The magnetic field is not in the direction of the position vector.
- Option B: The magnetic field is not in the direction of the current element.
- Option D: The magnetic field is perpendicular to the current element, but it also needs to be perpendicular to the position vector.

Key Concepts:
- Biot-Savart Law
- Cross Product of Vectors
- Magnetic Field due to Current Element

Memory Aids:
- Remember the right-hand rule for cross products.

Quick Tips:
- Visualize the current element and position vector to determine the direction of the magnetic field using the right-hand rule.

30 Hard

30 of 269Question permalink

Application +4 | -1

A charged particle moving along

+{x}

direction with velocity

{v}

enters a region where there is a uniform magnetic field

\vec{B}=-B \hat{k}

from

x=0

to

x=d

, the particle gets deflected at an angle

\theta

from its intial path. The specific charge of the particle is :

View answer and explanation

Correct option: D

\frac{{V} \sin \theta}{{Bd}}

Option D (

\frac{1}{3} \pi q f l^2

) is correct because the magnetic moment is calculated by integrating the current due to each infinitesimal charge element along the rod.

Solution approach

Correct Answer + Core Reason: $\frac{1}{3} \pi q f l^2$ . The magnetic moment is calculated by integrating the current due to each infinitesimal charge element.

Step-by-Step Solution:

Step 1: Consider a small element of length $dx$ at a distance $x$ from the pivot.

Step 2: The charge on this element is $dq = \frac{q}{l} dx$ .

Step 3: The current due to this charge element is $dI = dq \cdot f = \frac{q}{l} f dx$ .

Step 4: The area swept by this element is $dA = \pi x^2$ .

Step 5: The magnetic moment due to this element is $dM = dI \cdot dA = \frac{q}{l} f dx \cdot \pi x^2 = \frac{\pi q f}{l} x^2 dx$ .

Step 6: Integrate over the length of the rod to find the total magnetic moment.
$M = \int_0^l dM = \int_0^l \frac{\pi q f}{l} x^2 dx = \frac{\pi q f}{l} \int_0^l x^2 dx = \frac{\pi q f}{l} \left[\frac{x^3}{3}\right]_0^l = \frac{\pi q f}{l} \cdot \frac{l^3}{3} = \frac{1}{3} \pi q f l^2$

Why other options are wrong:
- Option A: zero: This is incorrect because the rotating charged rod does produce a magnetic moment.
- Option B: $\pi q f l^2$ : This is incorrect because the integration factor is missing.
- Option C: $\frac{1}{2} \pi q f l^2$ : This is incorrect because the integration factor is wrong.

Key Concepts:
- Magnetic moment.
- Integration.

Memory Aids:
- Remember to integrate over the entire length of the rod.

Quick Tips:
- Always consider a small element and then integrate to find the total quantity.

Your Progress

Magnetic effect of current practice · Mixed difficulty

Explore Other Chapters in PHYSICS

Key Takeaways

Continue Your Practice

Your practice