MAGNETISM AND MATTER MCQs for NEET UG PHYSICS — Easy

1 Easy

Analyzing +4 | -1

Assertion: If a unit north pole rotates around a current carrying wire then work has to be done.
Reason: Magnetic field produced by current is always non - conservative in nature.

View answer and explanation

Correct option: A

If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion

Option A is correct because the work done in moving a unit north pole around a current-carrying wire is non-zero, and this is due to the non-conservative nature of the magnetic field produced by the current.

Solution approach

Correct Answer + Core Reason: If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion. The magnetic field is non-conservative, hence work is done.

Step-by-Step Solution:

Step 1: Understand the concept of conservative and non-conservative fields.

Step 2: A conservative field is one where the work done in moving a test charge or mass between two points is independent of the path taken. Examples include gravitational and electrostatic fields.

Step 3: A non-conservative field is one where the work done depends on the path taken. Magnetic fields due to currents are non-conservative.

Step 4: When a unit north pole rotates around a current-carrying wire, the work done is given by the line integral of the magnetic force around the closed loop. Since the magnetic field is non-conservative, this integral is non-zero.

Step 5: Therefore, the assertion is true, and the reason correctly explains why the assertion is true.

Why other options are wrong:
- Option B: The reason is a correct explanation of the assertion.
- Option C: The reason is true.
- Option D: Both assertion and reason are true.

Key Concepts:
- Conservative and non-conservative fields
- Work done in a magnetic field

Memory Aids:
- Remember that magnetic fields due to currents are non-conservative.

Quick Tips:
- Understand the difference between conservative and non-conservative fields.

2 Easy

2 of 26Question permalink

Application +4 | -1

The magnetic length of a bar magnet is 15 cm then its geometric length is

View answer and explanation

Correct option: A

18 cm

Option A is correct because the geometric length is approximately 6/5 times the magnetic length for a bar magnet, resulting in 18 cm.

Solution approach

Correct Answer + Core Reason: 18 cm. The geometric length is approximately 6/5 times the magnetic length.

Step-by-Step Solution:

Step 1: Recall the relationship between magnetic length ( $L_m$ ) and geometric length ( $L_g$ ) for a bar magnet: $L_m \approx \frac{5}{6} L_g$ .

Step 2: Given the magnetic length $L_m = 15 \text{{ cm}}$ , we need to find the geometric length $L_g$ .

Step 3: Rearrange the formula to solve for $L_g$ : $L_g = \frac{6}{5} L_m$ .

Step 4: Substitute the given value: $L_g = \frac{6}{5} \times 15 \text{{ cm}}$ .

Step 5: Calculate the geometric length: $L_g = 18 \text{{ cm}}$ .

Why other options are wrong:
- Option B: 20 cm is not the correct geometric length based on the given magnetic length and the relationship between them.
- Option C: 25 cm is not the correct geometric length.
- Option D: 30 cm is not the correct geometric length.

Key Concepts:
- Magnetic length and geometric length of a bar magnet

Memory Aids:
- Remember the approximate ratio: $L_m \approx \frac{5}{6} L_g$ .

Quick Tips:
- Always use the correct relationship between magnetic and geometric lengths.

3 Easy

3 of 26Question permalink

Application +4 | -1

If area vector

\vec A = 2\vec i - 3\vec j + 6\vec k\,{m^2}

flux density vector

\vec B = - 3\vec i + \vec j + 2\vec k\,\left( {Wb/{m^2}} \right)

. Then magnetic flux linked with the coil is [Wb]

View answer and explanation

Correct option: B

3

Option B is correct because the magnetic flux is calculated by taking the dot product of the area vector and the magnetic flux density vector, which results in 3 Wb.

Solution approach

Correct Answer + Core Reason: 3. The magnetic flux is the dot product of the area vector and the magnetic flux density vector.

Step-by-Step Solution:

Step 1: Recall the formula for magnetic flux: $\Phi = \vec{B} \cdot \vec{A}$ .

Step 2: Given $\vec{A} = 2\vec{i} - 3\vec{j} + 6\vec{k} \, m^2$ and $\vec{B} = -3\vec{i} + \vec{j} + 2\vec{k} \, (Wb/m^2)$ .

Step 3: Calculate the dot product: $\Phi = (2\vec{i} - 3\vec{j} + 6\vec{k}) \cdot (-3\vec{i} + \vec{j} + 2\vec{k})$ .

Step 4: $\Phi = (2 \times -3) + (-3 \times 1) + (6 \times 2) = -6 - 3 + 12 = 3 \, Wb$ .

Why other options are wrong:
- Option A: 4 is not the correct magnetic flux based on the given vectors.
- Option C: 8 is not the correct magnetic flux.
- Option D: 12 is not the correct magnetic flux.

Key Concepts:
- Magnetic flux
- Dot product of vectors

Memory Aids:
- Remember that magnetic flux is a scalar quantity obtained from the dot product.

Quick Tips:
- Always use the correct formula for magnetic flux and perform the dot product accurately.

4 Easy

4 of 26Question permalink

Remembering +4 | -1

A bar magnet of moment

M

is placed in the magnetic field

B

. The torque acting on the magnet is

View answer and explanation

Correct option: A

\mathrm{M} \times \mathrm{B}

Option A (

\mathrm{M} \times \mathrm{B}

) is correct because the torque on a magnet in a magnetic field is given by the cross product of the magnetic moment and the magnetic field.

Solution approach

Correct Answer + Core Reason: $\mathrm{M} \times \mathrm{B}$ . The torque is the cross product of the magnetic moment and the magnetic field.

Step-by-Step Solution:

Step 1: Recall the formula for torque on a magnet in a magnetic field: $\vec \tau = \vec M \times \vec B$ .

Step 2: The formula represents the cross product of the magnetic moment $\vec M$ and the magnetic field $\vec B$ .

Why other options are wrong:
- **Option B ( $\mathrm{M}-\mathrm{B}$ ): This is incorrect because torque is not the difference between magnetic moment and magnetic field.
- Option C ( $\frac{1}{2} \mathrm{M} \times \mathrm{B}$ ): This is incorrect because the torque is not half the cross product of magnetic moment and magnetic field.
- Option D ( $\mathrm{M}+\mathrm{B}$ ): This is incorrect because torque is not the sum of magnetic moment and magnetic field.

Key Concepts:
- Torque on a magnetic dipole in a magnetic field.
- Cross product.

Memory Aids:**
- Remember the formula: $\vec \tau = \vec M \times \vec B$ .

Quick Tips:
- The torque is maximum when the magnetic moment and magnetic field are perpendicular.

5 Easy

5 of 26Question permalink

Application +4 | -1

When two magnetic moments are compared using equal distance method the deflections produced are

45^\circ

and

30^\circ

. If the length of magnets are in the ratio

1\;:2

, the ratio of their pole strength is

View answer and explanation

Correct option: D

2\sqrt 3 \;:1

Option D (

2\sqrt 3 \;:1

) is correct because it is derived from the tangent law and the given ratio of lengths.

Solution approach

Correct Answer + Core Reason: $2\sqrt 3 \;:1$ . The ratio of pole strengths is calculated using the tangent law and the given length ratio.

Step-by-Step Solution:

Step 1: Understand the equal distance method. The magnetic field due to a magnet at a point on its end-on position is given by $B = \frac{{\mu_0}}{{4\pi}} \cdot \frac{{2M}}{{{d^3}}}$ , where $M$ is the magnetic moment and $d$ is the distance.

Step 2: Relate magnetic moment to pole strength and length. The magnetic moment $M = m \cdot l$ , where $m$ is the pole strength and $l$ is the length of the magnet.

Step 3: Apply the tangent law. In the equal distance method, $\frac{{{B_1}}}{{{B_2}}} = \frac{{\tan {{\theta }_1}}}{{\tan {{\theta }_2}}}$ , where ${B_1}$ and ${B_2}$ are the magnetic fields due to the two magnets, and ${\theta _1}$ and ${\theta _2}$ are the deflections.

Step 4: Substitute the given values. ${\theta _1} = 45^\circ$ and ${\theta _2} = 30^\circ$ , so $\frac{{\tan 45^\circ}}{{\tan 30^\circ}} = \frac{1}{{\frac{1}{{\sqrt 3 }}}} = \sqrt 3$ .

Step 5: Relate the magnetic fields to the magnetic moments. $\frac{{{B_1}}}{{{B_2}}} = \frac{{\frac{{\mu_0}}{{4\pi}} \cdot \frac{{2{M_1}}}{{{d^3}}}}}{{\frac{{\mu_0}}{{4\pi}} \cdot \frac{{2{M_2}}}{{{d^3}}}}} = \frac{{{M_1}}}{{{M_2}}}$ .

Step 6: Relate the magnetic moments to the pole strengths and lengths. $\frac{{{M_1}}}{{{M_2}}} = \frac{{{m_1}{l_1}}}{{{m_2}{l_2}}}$ .

Step 7: Substitute the given length ratio. $\frac{{{l_1}}}{{{l_2}}} = \frac{1}{2}$ , so $\frac{{{m_1}}}{{{m_2}}} = \frac{{{M_1}}}{{{M_2}}} \cdot \frac{{{l_2}}}{{{l_1}}} = \sqrt 3 \cdot 2 = 2\sqrt 3$ .

Step 8: Final Answer: The ratio of their pole strength is $2\sqrt 3 \;:1$ .

Why other options are wrong:
- **Option A ( $3\;:1$ ): This is incorrect because it doesn't account for both the tangent of the angles and the length ratio correctly.
- Option B ( $3\;:2$ ): This is incorrect as it doesn't properly combine the tangent ratio and length ratio.
- Option C ( $\sqrt 3 \;:1$ ): This is incorrect because it only considers the tangent ratio of the angles and neglects the length ratio of the magnets.

Key Concepts:
- Tangent Law in Magnetism
- Magnetic Moment
- Pole Strength

Memory Aids:
- Tangent Law: The ratio of the magnetic fields is equal to the ratio of the tangents of the angles.

Quick Tips:**
- Remember to consider both the angle of deflection and the physical dimensions of the magnets when comparing magnetic moments.

6 Easy

6 of 26Question permalink

Remembering +4 | -1

The SI unit of Magnetic Permiability is:

View answer and explanation

Correct option: C

\mathrm{Hm}^{-1}

Option C (

\mathrm{Hm}^{-1}

) is correct because magnetic permeability is measured in Henry per meter, which is the SI unit.

Solution approach

Correct Answer + Core Reason: $\mathrm{Hm}^{-1}$ . Magnetic permeability is measured in Henry per meter.

Step-by-Step Solution:
Step 1: Recall the definition of magnetic permeability ( $\mu$ ). It relates the magnetic field strength (B) to the magnetizing field (H): $B = \mu H$ .
Step 2: Consider the units. B is in Tesla (T), and H is in Ampere per meter (A/m). Therefore, $\mu$ is in T/(A/m) = T.m/A.
Step 3: Recognize that 1 Henry (H) = 1 T.m²/A. Therefore, $\mu$ is also in H/m or $\mathrm{Hm}^{-1}$ .

Why other options are wrong:
- Option A: $\mathrm{Am}^{-1}$ is the unit for magnetizing field (H), not permeability.
- Option B: $\mathrm{Am}^{-2}$ is not a standard unit related to magnetic permeability.
- Option D: $\mathrm{Hm}^{-2}$ is not the correct unit for magnetic permeability.

Key Concepts:
- Magnetic permeability
- SI units

Memory Aids:
- Remember the relationship B = $\mu$ H and the units of B and H.

Quick Tips:
- Always check the units of physical quantities to ensure consistency.

7 Easy

7 of 26Question permalink

Understanding +4 | -1

A magnet is dropped down in infinitely long vertical
copper tube.

View answer and explanation

Correct option: A

The magnet moves with continuously increasing
velocity and ultimately acquires a constant
terminal velocity

Option A is correct. As the magnet falls, it induces a current in the copper tube, creating a magnetic field that opposes the magnet's motion, leading to a terminal velocity.

Solution approach

Correct Answer + Core Reason: The magnet moves with continuously increasing velocity and ultimately acquires a constant terminal velocity. This is due to the balance between gravitational force and the opposing force from induced currents.

Step-by-Step Solution:

Step 1: Lenz's Law: As the magnet falls through the copper tube, it induces a current in the tube due to the changing magnetic flux.

Step 2: Eddy Currents: These induced currents are called eddy currents and create their own magnetic field.

Step 3: Opposing Force: According to Lenz's Law, the induced magnetic field opposes the change in magnetic flux, creating an upward force on the magnet.

Step 4: Acceleration and Velocity: Initially, the gravitational force is greater than the opposing magnetic force, so the magnet accelerates downwards. As the velocity increases, the induced current and the opposing force also increase.

Step 5: Terminal Velocity: Eventually, the opposing magnetic force equals the gravitational force, resulting in zero net force and constant terminal velocity.

Why other options are wrong:
- Option B: The magnet does not come to rest; it reaches a terminal velocity.
- Option C: The acceleration is not constant; it decreases as the opposing force increases.
- Option D: The acceleration decreases as the velocity increases, eventually reaching zero when terminal velocity is achieved.

Key Concepts:
- Lenz's Law
- Eddy Currents
- Electromagnetic Induction
- Terminal Velocity

Memory Aids:
- Lenz's Law: "Nature abhors a change in flux."

Quick Tips:
- Remember that induced currents always oppose the change that caused them.

8 Easy

8 of 26Question permalink

Application +4 | -1

A magnet hung at

45^{\circ}

with magnetic meridian makes an angle of

60^{\circ}

with the horizontal. The actual value of the angle of dip is

View answer and explanation

Correct option: A

\tan ^{-1}\left(\sqrt{\frac{3}{2}}\right)

Option A (

\tan ^{-1}\left(\sqrt{\frac{3}{2}}\right)

) is correct because it's derived from the relationship between apparent and true angles of dip.

Solution approach

Correct Answer + Core Reason: $\tan ^{-1}\left(\sqrt{\frac{3}{2}}\right)$ . The actual angle of dip is calculated using the apparent angle and the angle with the magnetic meridian.

Step-by-Step Solution:

Step 1: Let $\theta$ be the true angle of dip and $\theta'$ be the apparent angle of dip. Let $\alpha$ be the angle the magnet makes with the magnetic meridian.

Step 2: We have $\tan{\theta'} = \frac{{\tan{\theta}}}{{\cos{\alpha}}}$ .

Step 3: Given $\alpha = 45^{\circ}$ and $\theta' = 60^{\circ}$ . So, $\tan{60^{\circ}} = \frac{{\tan{\theta}}}{{\cos{45^{\circ}}}}$ .

Step 4: $\sqrt{3} = \frac{{\tan{\theta}}}{{\frac{1}{{\sqrt{2}}}}} \implies \tan{\theta} = \sqrt{3} \times \frac{1}{{\sqrt{2}}} = \sqrt{\frac{3}{2}}$ .

Step 5: Therefore, $\theta = \tan^{-1}\left(\sqrt{\frac{3}{2}}\right)$ .

Why other options are wrong:
- Option B: $\tan ^{-1}(\sqrt{6})$ is incorrect. It's likely a result of incorrect manipulation of the trigonometric functions.
- Option C: $\tan ^{-1}\left(\sqrt{\frac{2}{3}}\right)$ is incorrect. It's the inverse of the correct answer.
- Option D: $\tan ^{-1}\left(\sqrt{\frac{1}{2}}\right)$ is incorrect. It's likely a result of incorrect manipulation of the trigonometric functions.

Key Concepts:
- Apparent and true angles of dip
- Relationship between them: $\tan{\theta'} = \frac{{\tan{\theta}}}{{\cos{\alpha}}}$

Memory Aids:
- Remember the formula relating true and apparent dip angles.

Quick Tips:
- Draw a diagram to visualize the angles.
- Be careful with trigonometric identities and manipulations.

9 Easy

9 of 26Question permalink

Understanding +4 | -1

A magnetic needle kept in a non-uniform
magnetic field, experiences

View answer and explanation

Correct option: C

both force and torque

Option C (both force and torque) is correct because a magnetic needle in a non-uniform magnetic field experiences both a net force and a torque.

Solution approach

Correct Answer + Core Reason: both force and torque. A non-uniform field exerts both a force and a torque.

Step-by-Step Solution:

Step 1: Consider a magnetic needle in a non-uniform magnetic field.

Step 2: In a non-uniform field, the forces on the north and south poles of the needle are unequal, resulting in a net force.

Step 3: Additionally, the non-uniform field can cause the magnetic needle to experience a torque, tending to align it with the field.

Step 4: Final Answer: both force and torque

Why other options are wrong:
- Option A: This is incorrect because a non-uniform field also exerts a torque.
- Option B: This is incorrect because a non-uniform field exerts both a force and a torque.
- Option D: This is incorrect because a non-uniform field also exerts a force.

Key Concepts:
- Magnetic force.
- Magnetic torque.
- Non-uniform magnetic field.

Memory Aids:
- Non-uniform field = both force and torque.

Quick Tips:
- Visualize the forces on the poles of the magnet in a non-uniform field.

10 Easy

10 of 26Question permalink

Application +4 | -1

The pole strength of a bar magnet is

0.08Am

and its length is

6cm

then magnetic moment of a bar magnet is

\left[ {A - {m^2}} \right]

View answer and explanation

Correct option: C

4.8 \times {10^{ - 3}}

Option C (

4.8 \times {10^{ - 3}}

) is correct because the magnetic moment is the product of pole strength and length.

Solution approach

Correct Answer + Core Reason: $4.8 \times {10^{ - 3}}$ . Magnetic moment is pole strength times length.

Step-by-Step Solution:
Step 1: Recall the formula for magnetic moment (M): $M = m \times l$ , where m is the pole strength and l is the length of the magnet.
Step 2: Convert the length from cm to meters: $l = 6 \text{{ cm}} = 0.06 \text{{ m}}$ .
Step 3: Substitute the given values into the formula: $M = 0.08 \text{{ Am}} \times 0.06 \text{{ m}}$ .
Step 4: Calculate the magnetic moment: $M = 0.0048 \text{{ A m}}^2 = 4.8 \times 10^{-3} \text{{ A m}}^2$ .

Why other options are wrong:
- Option A: $2 \times {10^{ - 3}}$ is incorrect.
- Option B: $8 \times {10^{ - 3}}$ is incorrect.
- Option D: $10 \times {10^{ - 3}}$ is incorrect.

Key Concepts:
- Magnetic moment
- Pole strength

Memory Aids:
- M = ml (Magnetic moment equals pole strength times length).

Quick Tips:
- Ensure consistent units (meters for length).

11 Easy

11 of 26Question permalink

Understanding +4 | -1

Assertion (A) : The magnetic field lines are
continuous and form closed loops.
Reason (R) : Magnetic monopole does not exist.
The correct option among the following is

View answer and explanation

Correct option: A

(A) is true and (R) is the correct explanation
for (A)

Option A ((A) is true and (R) is the correct explanation for (A)) is correct because magnetic field lines are continuous and form closed loops due to the non-existence of magnetic monopoles.

Solution approach

Correct Answer + Core Reason: (A) is true and (R) is the correct explanation for (A). Magnetic field lines form closed loops because magnetic monopoles do not exist.

Step-by-Step Solution:

Step 1: Recall the properties of magnetic field lines.

Step 2: Magnetic field lines are continuous and form closed loops.

Step 3: The reason for this is that magnetic monopoles (isolated north or south poles) do not exist. Magnetic field lines always originate from a north pole and terminate at a south pole, forming a closed loop.

Why other options are wrong:
- Option B: (A) is true. (R ) is not the correct explanation of (A): This is incorrect because the non-existence of magnetic monopoles is the reason for the closed loop nature of magnetic field lines.
- Option C: (A) is true but (R) is false: This is incorrect because both the assertion and the reason are true.
- Option D: (A) is false but (R) is true: This is incorrect because the assertion is also true.

Key Concepts:
- Magnetic field lines
- Magnetic monopoles

Memory Aids:
- Remember that magnetic field lines always form closed loops because isolated magnetic poles do not exist.

Quick Tips:
- The absence of magnetic monopoles is a fundamental concept in magnetism.

12 Easy

12 of 26Question permalink

Understanding +4 | -1

The dimensions of B is

View answer and explanation

Correct option: A

\left[ {{M^1}{T^{ - 2}}{A^{ - 1}}} \right]

Option A (

\left[ {{M^1}{T^{ - 2}}{A^{ - 1}}} \right]

) is correct because it represents the correct dimensions of magnetic field B, derived from the Lorentz force equation.

Solution approach

Correct Answer + Core Reason: $\left[ {{M^1}{T^{ - 2}}{A^{ - 1}}} \right]$ . This is the correct dimension for magnetic field B.

Step-by-Step Solution:

Step 1: Recall the Lorentz force equation: $F = qvB$ , where F is force, q is charge, v is velocity, and B is the magnetic field.

Step 2: Rearrange the equation to solve for B: $B = \frac{F}{qv}$ .

Step 3: Determine the dimensions of each quantity: [F] = $MLT^{-2}$ , [q] = AT, [v] = $LT^{-1}$ .

Step 4: Substitute the dimensions into the equation for B: $[B] = \frac{MLT^{-2}}{(AT)(LT^{-1})} = \frac{MLT^{-2}}{AL} = MT^{-2}A^{-1}$ .

Why other options are wrong:
- Option B: $\left[ {{M^{ - 1}}{T^{ - 2}}{A^{ - 1}}} \right]$ : This is incorrect as the mass dimension should be positive.
- Option C: $\left[ {{M^0}{L^0}{T^0}} \right]$ : This is incorrect as the magnetic field has dimensions.
- Option D: ${L^2}$ : This is incorrect as it only represents an area dimension.

Key Concepts:
- Lorentz force
- Dimensional analysis

Memory Aids:
- Remember the Lorentz force equation: $F = qvB$ .

Quick Tips:
- Use the Lorentz force equation to derive the dimensions of the magnetic field.

13 Easy

13 of 26Question permalink

Application +4 | -1

A bar magnet has a magnetic moment

2.5J{T^{ - 1}}

and is placed in a magnetic field of

0.2T

. Calculate the work done in turning the magnet from parallel to antiparallel position relative to field direction is

View answer and explanation

Correct option: B

1J

Option B is correct. The work done is calculated using the formula

W = MB(\cos{\theta_1} - \cos{\theta_2})

, where θ₁ = 0° and θ₂ = 180°.

Solution approach

Correct Answer + Core Reason: 1J. The work done is calculated using the potential energy difference between the parallel and antiparallel positions.

Step-by-Step Solution:

Step 1: Recall the formula for the potential energy of a magnetic dipole in a magnetic field: $U = -MB \cos{\theta}$

Step 2: The work done in turning the magnet from parallel ( $\theta_1 = 0^\circ$ ) to antiparallel ( $\theta_2 = 180^\circ$ ) is given by: $W = U_2 - U_1 = -MB \cos{\theta_2} - (-MB \cos{\theta_1}) = MB(\cos{\theta_1} - \cos{\theta_2})$

Step 3: Substitute the given values: $M = 2.5 \text{ JT}^{-1}$ , $B = 0.2 \text{ T}$ , $\theta_1 = 0^\circ$ , $\theta_2 = 180^\circ$

Step 4: Calculate the work done: $W = 2.5 \times 0.2 (\cos{0^\circ} - \cos{180^\circ}) = 2.5 \times 0.2 (1 - (-1)) = 2.5 \times 0.2 \times 2 = 1 \text{ J}$

Why other options are wrong:
- Option A: Incorrect due to incorrect calculation.
- Option C: Incorrect due to incorrect calculation.
- Option D: Incorrect due to incorrect calculation.

Key Concepts:
- Potential energy of a magnetic dipole
- Work done

Memory Aids:
- $U = -MB \cos{\theta}$

Quick Tips:
- Remember the sign convention for potential energy.

14 Easy

14 of 26Question permalink

Application +4 | -1

The vertical component of the earth's magnetic field is

6 \times 10^{-5} \mathrm{~T}

at any place where the angle of dip is

37^{\circ}

. The earth's resultant magnetic field at that place will be

\left(\right.

Given

\left.\tan 37^{\circ}=\frac{3}{4}\right)

View answer and explanation

Correct option: D

1 \times 10^{-4} \mathrm{~T}

Option D (

1 \times 10^{-4} \mathrm{~T}

) is correct because it's calculated using the relationship between the vertical component, angle of dip, and the earth's resultant magnetic field.

Solution approach

Correct Answer + Core Reason: $1 \times 10^{-4} \mathrm{~T}$ . The earth's magnetic field is calculated using the vertical component and the angle of dip.

Step-by-Step Solution:

Step 1: Given: Vertical component $B_V = 6 \times 10^{-5} \mathrm{~T}$ and angle of dip $\delta = 37^{\circ}$ . We need to find the earth's resultant magnetic field $B$ .

Step 2: We know that $B_V = B \sin{\delta}$ . Therefore, $B = \frac{{B_V}}{{\sin{\delta}}}$ .

Step 3: Since $\tan{37^{\circ}} = \frac{3}{4}$ , we can find $\sin{37^{\circ}}$ using a right triangle. If the opposite side is 3 and the adjacent side is 4, the hypotenuse is 5. So, $\sin{37^{\circ}} = \frac{3}{5} = 0.6$ .

Step 4: Substitute the values: $B = \frac{{6 \times 10^{-5}}}{{0.6}} = 10 \times 10^{-5} = 1 \times 10^{-4} \mathrm{~T}$ .

Why other options are wrong:
- Option A: $8 \times 10^{-5} \mathrm{~T}$ is incorrect. It might be a result of using $\cos$ instead of $\sin$ or incorrect calculation.
- Option B: $6 \times 10^{-5} \mathrm{~T}$ is just the vertical component, not the resultant field.
- Option C: $5 \times 10^{-4} \mathrm{~T}$ is incorrect. It's likely a calculation error.

Key Concepts:
- Earth's magnetic field components
- Angle of dip
- Relationship between vertical component, horizontal component, and resultant magnetic field: $B_V = B \sin{\delta}$

Memory Aids:
- Remember that the vertical component involves $\sin{\delta}$ .

Quick Tips:
- Always draw a diagram to visualize the components of the magnetic field.
- Use the given $\tan{\delta}$ to find $\sin{\delta}$ and $\cos{\delta}$ .

15 Easy

15 of 26Question permalink

Understanding +4 | -1

The area enclosed by a hysteresis loop is a measure of

View answer and explanation

Correct option: D

Energy loss per cycle

Option D (Energy loss per cycle) is correct because the area enclosed by the hysteresis loop represents the energy dissipated as heat during each magnetization-demagnetization cycle.

Solution approach

Correct Answer + Core Reason: Energy loss per cycle. The area enclosed by the hysteresis loop represents the energy dissipated as heat.

Step-by-Step Solution:

Step 1: Define hysteresis loop. A hysteresis loop is a graph showing the relationship between the magnetic field strength (H) and the magnetization (B) of a ferromagnetic material.

Step 2: Explain the area enclosed by the loop.\nThe area enclosed by the hysteresis loop represents the energy loss per cycle due to the irreversible processes of magnetization and demagnetization. This energy is dissipated as heat.

Step 3: Final Answer: Energy loss per cycle

Why other options are wrong:
- Option A: Retentivity is the residual magnetization when the applied field is zero, represented by the point where the loop intersects the B-axis.
- Option B: Susceptibility is the degree of magnetization of a material in response to an applied magnetic field, related to the slope of the initial magnetization curve.
- Option C: Permeability is the measure of the ability of a material to support the formation of magnetic fields, also related to the slope of the magnetization curve.

Key Concepts:
- Hysteresis loop
- Energy loss

Memory Aids:
- Hysteresis loop area = energy loss per cycle.

Quick Tips:
- Remember what the hysteresis loop represents.

16 Easy

16 of 26Question permalink

Remembering +4 | -1

The magnet can be completely demagnetized by

View answer and explanation

Correct option: D

A reverse field of appropriate strength

Option D (A reverse field of appropriate strength) is correct because applying a reverse magnetic field of sufficient strength can randomize the magnetic domains, leading to complete demagnetization.

Solution approach

Correct Answer + Core Reason: A reverse field of appropriate strength. Applying a reverse magnetic field can randomize the magnetic domains.

Step-by-Step Solution:

Step 1: Understand demagnetization. Demagnetization is the process of removing the magnetic properties of a material.

Step 2: Analyze the options.
- Breaking the magnet into small pieces does not completely demagnetize each piece.
- Heating it slightly might reduce the magnetization but not completely demagnetize it.
- Dropping it into ice-cold water has little to no effect on demagnetization.
- Applying a reverse field of appropriate strength can randomize the magnetic domains, leading to complete demagnetization.

Step 3: Final Answer: A reverse field of appropriate strength

Why other options are wrong:
- Option A: Breaking the magnet into small pieces only reduces the overall magnetic effect, but each piece remains magnetized.
- Option B: Heating slightly may reduce magnetization, but it doesn't guarantee complete demagnetization.
- Option C: Dropping it into ice-cold water has minimal impact on the magnetic properties.

Key Concepts:
- Demagnetization
- Magnetic domains
- Hysteresis

Memory Aids:
- Applying a reverse field is an effective way to demagnetize a material.

Quick Tips:
- Consider the methods used to demagnetize materials.

17 Easy

17 of 26Question permalink

Application +4 | -1

A magnet is suspended in such a way that it oscillates in the horizontal plane. It makes

20\;oscillations\;per\;minute

at a place where dip angle is

30^\circ

and

15\;oscillations\;per

minute

at a place where dip angle

60^\circ

. The ratio of total earth’s magnetic field at the two places is

View answer and explanation

Correct option: B

16\;:9\sqrt 3

Option B is correct because in equilibrium, the torques due to both magnetic fields must be equal and opposite, leading to the given relationship.

Solution approach

Correct Answer + Core Reason: $B\sin \mathrm{\theta } = H\cos \mathrm{\theta }$ . The torques due to the two fields balance each other.

Step-by-Step Solution:

Step 1: The torque on a magnetic dipole in a magnetic field is given by:
$\tau = M \times B = MB\sin{\theta}$ \nwhere $M$ is the magnetic moment, $B$ is the magnetic field, and $\theta$ is the angle between the magnetic moment and the magnetic field.

Step 2: In this case, the magnetic dipole is placed in two perpendicular magnetic fields, $B$ and $H$ . The dipole makes an angle $\theta$ with $B$ . Therefore, the angle with $H$ is $90^\circ - \theta$ .

Step 3: The torque due to field $B$ is $\tau_B = MB\sin{\theta}$ .

Step 4: The torque due to field $H$ is $\tau_H = MH\sin{(90^\circ - \theta)} = MH\cos{\theta}$ .

Step 5: For the dipole to be in equilibrium, the torques must balance each other:
$\tau_B = \tau_H$
$MB\sin{\theta} = MH\cos{\theta}$

Step 6: Divide both sides by $M$ :
$B\sin{\theta} = H\cos{\theta}$

Why other options are wrong:
- Option A: This is incorrect because it only holds true when $\theta = 45^\circ$ .
- Option C: This is incorrect because the sine and cosine terms are swapped.
- Option D: This is incorrect because it implies $B = H\tan{\theta}$ , which is the inverse of the correct relationship.

Key Concepts:
- Torque on a magnetic dipole in a magnetic field
- Equilibrium of torques

Memory Aids:
- Remember the torque formula: $\tau = MB\sin{\theta}$

Quick Tips:
- Always consider the angles between the magnetic moment and the magnetic fields.

18 Easy

18 of 26Question permalink

Application +4 | -1

Two magnets of magnetic moments

4\,A{m^2}

and

3\,A{m^2}

are joined to form a cross

\left( + \right)

then the magnetic moment of the combination is

\left[ {A{m^2}} \right]

View answer and explanation

Correct option: A

5

Option A (5) is correct because the magnetic moments are perpendicular, so the resultant is found using the Pythagorean theorem.

Solution approach

Correct Answer + Core Reason: 5. The magnetic moments are perpendicular, use Pythagorean theorem.

Step-by-Step Solution:
Step 1: Since the magnets are joined to form a cross, their magnetic moments are perpendicular to each other.
Step 2: The resultant magnetic moment can be found using the Pythagorean theorem: $M_{resultant} = \sqrt{{M_1^2 + M_2^2}}$ .
Step 3: Substitute the given values: $M_{resultant} = \sqrt{{(4)^2 + (3)^2}} = \sqrt{{16 + 9}} = \sqrt{{25}}$ .
Step 4: Calculate the resultant magnetic moment: $M_{resultant} = 5 \text{{ A m}}^2$ .

Why other options are wrong:
- Option B: 6 is incorrect.
- Option C: 8 is incorrect.
- Option D: 10 is incorrect.

Key Concepts:
- Magnetic moment
- Vector addition

Memory Aids:
- Perpendicular vectors: Use Pythagorean theorem.

Quick Tips:
- Remember to treat magnetic moments as vectors.

19 Easy

19 of 26Question permalink

Understanding +4 | -1

When a ferromagnetic substance is heated to a temperature above its Curie temperature, it:

View answer and explanation

Correct option: A

Behaves like a paramagnetic substance

Option A is correct. Above the Curie temperature, the thermal energy disrupts the alignment of magnetic domains, causing the ferromagnetic substance to behave like a paramagnetic substance.

Solution approach

Correct Answer + Core Reason: Behaves like a paramagnetic substance. Above the Curie temperature, the substance loses its ferromagnetic properties and becomes paramagnetic.

Step-by-Step Solution:

Step 1: Define Curie temperature: The Curie temperature is the temperature above which certain materials lose their permanent magnetic properties, to be replaced by induced magnetism.

Step 2: Ferromagnetic materials: Below the Curie temperature, ferromagnetic materials exhibit spontaneous magnetization due to the alignment of magnetic domains.

Step 3: Paramagnetic materials: Paramagnetic materials exhibit weak attraction to magnetic fields due to the partial alignment of atomic magnetic dipoles.

Step 4: Above Curie temperature: When a ferromagnetic material is heated above its Curie temperature, the thermal energy disrupts the alignment of magnetic domains, causing it to behave like a paramagnetic material.

Why other options are wrong:
- Option B: Diamagnetic substances are weakly repelled by magnetic fields, which is not the behavior observed above the Curie temperature.
- Option C: Ferromagnetism is lost above the Curie temperature due to the randomization of magnetic domains.
- Option D: Permanent magnetization is a characteristic of ferromagnetism, which is lost above the Curie temperature.

Key Concepts:
- Ferromagnetism
- Paramagnetism
- Curie Temperature

Memory Aids:
- Curie temperature: Think of it as the "breaking point" for ferromagnetism.

Quick Tips:
- Remember the order: Ferromagnetic → Paramagnetic → Diamagnetic (with increasing temperature or disorder).

20 Easy

20 of 26Question permalink

Application +4 | -1

At a certain place the angle of dip is

30^{\circ}

and the horizontal component of earth's magnetic field is

0.5 \,\mathrm{G}

. The earth's total magnetic field (in G), at that certain place, is :

View answer and explanation

Correct option: A

\frac{1}{\sqrt{3}}

Option A (

\frac{1}{\sqrt{3}}

) is correct because it represents the total magnetic field calculated from the horizontal component and the angle of dip.

Solution approach

Correct Answer + Core Reason: $\frac{1}{\sqrt{3}}$ . The earth's total magnetic field is calculated using the horizontal component and the angle of dip.

Step-by-Step Solution:

Step 1: Identify the given information. Angle of dip, $\delta = 30^\circ$ , and horizontal component of Earth's magnetic field, $B_H = 0.5 \text{ G}$ .

Step 2: Recall the relationship between the horizontal component, total magnetic field, and angle of dip. $B_H = B \cos \delta$ , where $B$ is the total magnetic field.

Step 3: Rearrange the formula to solve for the total magnetic field. $B = \frac{{B_H}}{{\cos \delta}}$ .

Step 4: Substitute the given values. $B = \frac{{0.5}}{{\cos 30^\circ}}$ .

Step 5: Calculate the value. $\cos 30^\circ = \frac{{\sqrt 3 }}{2}$ , so $B = \frac{{0.5}}{{\frac{{\sqrt 3 }}{2}}} = \frac{1}{{\sqrt 3 }} \text{ G}$ .

Why other options are wrong:
- **Option B ( $\frac{1}{2}$ ): This is incorrect because it does not correctly use the cosine of the angle of dip in the formula.
- Option C ( $\sqrt{3}$ ): This is incorrect because it inverts the relationship between the horizontal component and the total magnetic field.
- Option D ( $1$ ): This is incorrect because it doesn't account for the angle of dip when calculating the total magnetic field.

Key Concepts:
- Angle of Dip
- Earth's Magnetic Field
- Horizontal Component of Earth's Magnetic Field

Memory Aids:**
- $B_H = B \cos \delta$ (Horizontal component is related to total field by cosine of dip angle)

Quick Tips:
- Remember the relationship between the horizontal component, vertical component, total magnetic field, and angle of dip.

21 Easy

21 of 26Question permalink

Understanding +4 | -1

The magnetic induction and magnetising field intensity in a sample of magnetic material are

\mathrm{B}

and

\mathrm{H}

respectively. The magnetic susceptibility of the material is:

View answer and explanation

Correct option: B

\frac{\mathrm{B}}{\mu_0 \mathrm{H}}-1

Option B (

\frac{\mathrm{B}}{\mu_0 \mathrm{H}}-1

) is correct because it represents the formula for magnetic susceptibility,

\chi_m = \frac{B}{\mu_0 H} - 1

.

Solution approach

Correct Answer + Core Reason: $\frac{\mathrm{B}}{\mu_0 \mathrm{H}}-1$ . This is the formula for magnetic susceptibility.

Step-by-Step Solution:

Step 1: Recall the relationship between magnetic induction (B), magnetizing field (H), and magnetization (I): $B = \mu_0(H + I)$ .

Step 2: Recall the definition of magnetic susceptibility ( $\chi_m$ ): $I = \chi_m H$ .

Step 3: Substitute $I = \chi_m H$ into the equation $B = \mu_0(H + I)$ : $B = \mu_0(H + \chi_m H)$ .

Step 4: Solve for $\chi_m$ : $B = \mu_0 H(1 + \chi_m) \implies \frac{B}{\mu_0 H} = 1 + \chi_m \implies \chi_m = \frac{B}{\mu_0 H} - 1$ .

Why other options are wrong:
- Option A: $\frac{\mathrm{B}}{\mu_0 \mathrm{H}}$ : This is incorrect as it doesn't subtract 1.
- Option C: $\frac{\mathrm{B}}{\mathrm{H}}-1$ : This is incorrect as it doesn't include $\mu_0$ in the denominator.
- Option D: $\frac{\mu_0 \mathrm{H}}{\mathrm{B}}-1$ : This is incorrect as it is the inverse of the correct expression minus 1.

Key Concepts:
- Magnetic susceptibility
- Magnetic induction
- Magnetizing field intensity

Memory Aids:
- Remember the formula $\chi_m = \frac{B}{\mu_0 H} - 1$ .

Quick Tips:
- Ensure you know the relationship between B, H, I, and $\chi_m$ .

22 Easy

22 of 26Question permalink

Application +4 | -1

The period of oscillation of a freely suspended bar magnet is 4 s. If it is cut into two equal parts in length, then the time period of each part will be

View answer and explanation

Correct option: B

2 s

Option B is correct because the current required to produce a deflection in a tangent galvanometer is calculated using the tangent law, and the given values lead to approximately 1.1 A.

Solution approach

Correct Answer + Core Reason: 1.1 A. The current is calculated using the tangent galvanometer formula.

Step-by-Step Solution:

Step 1: The formula for the tangent galvanometer is:
$B = B_H \tan{\theta} = \frac{\mu_0 n I}{2r}$ \nwhere $B$ is the magnetic field produced by the coil, $B_H$ is the horizontal component of the Earth's magnetic field, $\theta$ is the deflection angle, $\mu_0$ is the permeability of free space ( $4\pi \times 10^{-7} \text{{ T m/A}}$ ), $n$ is the number of turns, $I$ is the current, and $r$ is the radius of the coil.

Step 2: We are given: $n = 10$ , $r = 0.1 \text{{ m}}$ , $\theta = 60^\circ$ , and $B_H = 4 \times 10^{-5} \text{{ T}}$ .

Step 3: Rearrange the formula to solve for $I$ :
$I = \frac{2rB_H \tan{\theta}}{\mu_0 n}$

Step 4: Substitute the given values:
$I = \frac{2 \times 0.1 \times 4 \times 10^{-5} \times \tan{60^\circ}}{4\pi \times 10^{-7} \times 10} = \frac{8 \times 10^{-6} \times \sqrt{3}}{4\pi \times 10^{-6}} = \frac{2\sqrt{3}}{\pi}$

Step 5: Calculate the value:
$I = \frac{2 \times 1.732}{3.14} \approx \frac{3.464}{3.14} \approx 1.1 \text{{ A}}$

Why other options are wrong:
- Option A: This is incorrect because it results from an incorrect calculation or using the wrong formula.
- Option C: This is incorrect because it results from an incorrect calculation or using the wrong formula.
- Option D: This is incorrect because it results from an incorrect calculation or using the wrong formula.

Key Concepts:
- Tangent galvanometer
- Tangent law
- Magnetic field due to a current-carrying coil

Memory Aids:
- Remember the tangent galvanometer formula: $B = B_H \tan{\theta} = \frac{\mu_0 n I}{2r}$

Quick Tips:
- Ensure that all units are consistent before substituting into the formula.

23 Easy

23 of 26Question permalink

Application +4 | -1

A magnet of magnetic moment

6 \mathrm{JT}^{-1}

is aligned in the direction of magnetic field of

0.3 \mathrm{~T}

. The net work done to bring the magnet normal to the magnetic field is

View answer and explanation

Correct option: C

1.8 \mathrm{~J}

Option C is correct because the work done is calculated using the formula W = MB(cosθ₁ - cosθ₂), where θ₁ = 0° and θ₂ = 90°, resulting in 1.8 J.

Solution approach

Correct Answer + Core Reason: $1.8 \mathrm{~J}$ . The work done is calculated using $W = MB(\cos{\theta_1} - \cos{\theta_2})$ .

Step-by-Step Solution:

Step 1: The work done to bring the magnet normal to the magnetic field is given by $W = MB(\cos{\theta_1} - \cos{\theta_2})$ , where $M$ is the magnetic moment, $B$ is the magnetic field, $\theta_1$ is the initial angle, and $\theta_2$ is the final angle.

Step 2: Given: $M = 6 \,JT^{-1}$ , $B = 0.3 \,T$ . Initially, the magnet is aligned with the field, so $\theta_1 = 0^\circ$ . The magnet is brought normal to the field, so $\theta_2 = 90^\circ$ .

Step 3: Substitute the values into the work done formula: $W = 6 \times 0.3 \times (\cos{0^\circ} - \cos{90^\circ}) = 6 \times 0.3 \times (1 - 0) = 1.8 \,J$ .

Why other options are wrong:
- Option A: $2 \mathrm{~J}$ : Incorrect calculation.
- Option B: $1 \mathrm{~J}$ : Incorrect calculation.
- Option D: $2.4 \mathrm{~J}$ : Incorrect calculation.

Key Concepts:
- Work done in rotating a magnet in a magnetic field: $W = MB(\cos{\theta_1} - \cos{\theta_2})$

Memory Aids:
- Remember the formula for work done.

Quick Tips:
- Ensure the angles are in degrees when using the cosine function.

24 Easy

24 of 26Question permalink

Application +4 | -1

Two electrons move parallel to each other with equal speed

V.

The ratio of magnetic and electric force between them is

View answer and explanation

Correct option: A

\frac{{{V^2}}}{{{C^2}}}

Option A (

\frac{{{V^2}}}{{{C^2}}}

) is correct because the ratio of magnetic to electric force is proportional to the square of the velocity divided by the square of the speed of light.

Solution approach

Correct Answer + Core Reason: $\frac{{{V^2}}}{{{C^2}}}$ . The ratio of magnetic force to electric force is proportional to $(v/c)^2$ .

Step-by-Step Solution:

Step 1: Electric force between two electrons: $F_e = \frac{1}{{4\pi \epsilon_0}} \frac{{e^2}}{{r^2}}$

Step 2: Magnetic force between two parallel moving electrons: $F_m = \frac{{\mu_0}}{{4\pi}} \frac{{e^2 v^2}}{{r^2}}$

Step 3: Ratio of magnetic force to electric force: $\frac{{F_m}}{{F_e}} = \frac{{\frac{{\mu_0}}{{4\pi}} \frac{{e^2 v^2}}{{r^2}}}}{{\frac{1}{{4\pi \epsilon_0}} \frac{{e^2}}{{r^2}}}} = \mu_0 \epsilon_0 v^2$

Step 4: Since $c^2 = \frac{1}{{\mu_0 \epsilon_0}}$ , we have $\mu_0 \epsilon_0 = \frac{1}{{c^2}}$

Step 5: Therefore, $\frac{{F_m}}{{F_e}} = \frac{{v^2}}{{c^2}}$

Why other options are wrong:
- **Option B ( $\frac{V}{C}$ ): This is incorrect because the ratio involves the square of the velocity divided by the square of the speed of light.
- Option C ( $\frac{{{C^2}}}{{{V^2}}}$ ): This is the inverse of the correct ratio.
- Option D ( $\frac{C}{V}$ ): This is the inverse of the square root of the correct ratio.

Key Concepts:
- Electric force
- Magnetic force
- Speed of light

Memory Aids:
- Remember the relationship between electric and magnetic forces for moving charges.

Quick Tips:**
- Recall the formulas for electric and magnetic forces between moving charges.

25 Easy

25 of 26Question permalink

Application +4 | -1

A bar magnet of length

10cm

and pole strength

2\,Am

makes an angle

60^\circ

with a uniform magnetic field of induction

50T

. The couple acting on it is

View answer and explanation

Correct option: A

5\sqrt 3 Nm

Option A is correct because the torque is calculated using the formula τ = MBLsinθ, where L is the length, and substituting the given values yields 5√3 Nm.

Solution approach

Correct Answer + Core Reason: $5\sqrt 3 Nm$ . The torque is calculated using $\tau = MBL\sin{\theta}$ .

Step-by-Step Solution:

Step 1: The torque on a bar magnet in a uniform magnetic field is given by $\tau = M B \sin{\theta}$ , where $M$ is the magnetic moment, $B$ is the magnetic field, and $\theta$ is the angle between the magnetic moment and the magnetic field.

Step 2: The magnetic moment $M$ is given by $M = m \times l$ , where $m$ is the pole strength and $l$ is the length of the magnet. Here, $m = 2 \,Am$ and $l = 10cm = 0.1m$ , so $M = 2 \times 0.1 = 0.2 \,Am^2$ .

Step 3: The magnetic field is $B = 50 \,T$ , and the angle is $\theta = 60^\circ$ .

Step 4: Substitute the values into the torque formula: $\tau = 0.2 \times 50 \times \sin{60^\circ} = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3} \,Nm$ .

Why other options are wrong:
- Option B: $\sqrt 3 Nm$ : Incorrect calculation.
- Option C: $10\sqrt 3 Nm$ : Incorrect calculation.
- Option D: $20\sqrt 3 Nm$ : Incorrect calculation.

Key Concepts:
- Torque on a magnet in a magnetic field: $\tau = MB\sin{\theta}$
- Magnetic moment: $M = m \times l$

Memory Aids:
- Remember the formula for torque and magnetic moment.

Quick Tips:
- Convert all units to SI units before calculating.

26 Easy

26 of 26Question permalink

Application +4 | -1

A bar magnet of magnetic moment

2A{m^2}

is free to rotate about a vertical axis passing through its centre. The magnet is released from rest from east – west position. Then the K.E of the magnet as it takes N – S position is

\left[ {{B_H} = 25\mu T} \right]

View answer and explanation

Correct option: B

50\mu J

Option B (

50\mu J

) is correct because the kinetic energy gained equals the potential energy lost as the magnet aligns with the field.

Solution approach

Correct Answer + Core Reason: $50\mu J$ . Kinetic energy is equal to the potential energy difference between east-west and north-south positions.

Step-by-Step Solution:

Step 1: The potential energy of a magnet in a magnetic field is given by $U = -MB_H \cos{\theta}$ , where $M$ is the magnetic moment, $B_H$ is the horizontal component of the Earth's magnetic field, and $\theta$ is the angle between the magnetic moment and the field.

Step 2: Initially, the magnet is in the east-west position, so $\theta_1 = 90^{\circ}$ . Finally, the magnet is in the north-south position, so $\theta_2 = 0^{\circ}$ .

Step 3: The change in potential energy is $\Delta U = U_2 - U_1 = -MB_H \cos{0^{\circ}} - (-MB_H \cos{90^{\circ}}) = -MB_H(1) - (-MB_H(0)) = -MB_H$ .

Step 4: The kinetic energy gained is equal to the negative of the change in potential energy: $KE = -\Delta U = MB_H$ .

Step 5: Given $M = 2 \text{ Am}^2$ and $B_H = 25 \mu T = 25 \times 10^{-6} \text{ T}$ .

Step 6: $KE = 2 \times 25 \times 10^{-6} = 50 \times 10^{-6} \text{ J} = 50 \mu J$ .

Why other options are wrong:
- Option A: $25\mu J$ is incorrect. It's likely a result of using only M or B in the calculation.
- Option C: $100\mu J$ is incorrect. It's likely a calculation error.
- Option D: $1.25\mu J$ is incorrect. It's likely a result of incorrect units or a miscalculation.

Key Concepts:
- Potential energy of a magnet in a magnetic field: $U = -MB\cos{\theta}$
- Conservation of energy

Memory Aids:
- Remember the potential energy formula and the relationship between potential and kinetic energy.

Quick Tips:
- Pay attention to the units and convert them if necessary.
- Remember that kinetic energy gained is equal to the potential energy lost.

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