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JEE • PHYSICS • MAGNETIC EFFECT OF CURRENT

Magnetic effect of current practice · Mixed difficulty

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Showing 271-300 of 418 questions

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271 Medium
271 of 418
Understanding +4 | -1
In the cyclotron, as radius of the circular path of the charged particle increases :
(ω=(\omega= angular velocity, v=v= linear velocity)
View answer and explanation
Correct option: C
vv increases, ω\omega remains constant
Option C is correct because in a cyclotron, as the radius increases, the velocity increases, but the angular velocity remains constant.
272 Medium
272 of 418
Understanding +4 | -1
A coil carrying electric current is placed in
uniform magnetic field, then
View answer and explanation
Correct option: A
torque is formed
Option A is correct. A current-carrying coil in a uniform magnetic field experiences a torque due to the magnetic forces on the current elements.
273 Easy
273 of 418
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 carried by EM fields, and Ampere's law is derived from Biot-Savart's law.
274 Medium
274 of 418
Application +4 | -1
A tangent galvanometer shows a deflection 45∘45^\circ when 10  mA10\;mA current passes through it. If the horizontal component of the earth’s field is 3.6×10−5T3.6 \times {10^{ - 5}}T  and radius of the coil is 10  cm10\;cm. The number of turns in the coil is
View answer and explanation
Correct option: C
570
Option C (570) is correct because the number of turns is calculated using the tangent galvanometer formula, considering the deflection, current, earth's magnetic field, and coil radius.
275 Medium
275 of 418
Understanding +4 | -1
A stray magnetic field does ‘NOT’ affect the
deflection of moving coil galvanometer because
View answer and explanation
Correct option: B
magnetic field inside galvanometer is very strong
Option B is correct because the strong magnetic field inside the galvanometer dominates any stray magnetic fields, minimizing their effect on the coil's deflection.
276 Medium
276 of 418
Analyzing +4 | -1
A charged particle moves in a gravity free space without change in velocity. Which of the following is not possible in that space?
View answer and explanation
Correct option: B
E≠0,B=0E \neq 0, B=0
Option B (E≠0,B=0E \neq 0, B=0) is correct because a charged particle cannot move with constant velocity in a gravity-free space if only an electric field is present.
277 Easy
277 of 418
Application +4 | -1
If 2%2 \% of the main current is to be passed through the galvanometer of resistance GG, the resistance of shunt required is
View answer and explanation
Correct option: A
G49\frac{G}{49}
Option A (G49\frac{G}{49}) is correct because it is calculated using the formula for shunt resistance S=IGI−IGGS = \frac{I_G}{I - I_G}G.
278 Medium
278 of 418
Application +4 | -1
A wire is formed into a circle having a diameter of 10.0 cm10.0 \mathrm{~cm} and placed in a uniform magnetic field of 3.00 mT3.00 \mathrm{~mT}. The wire carries a current of 5.00 A5.00 \mathrm{~A}. Find (a) the maximum torque on the wire, and (b) the range of potential energies of the wire-field system for different orientations of the circle.
View answer and explanation
Correct option: A
375π×10−7 Nm,+MB,−MB375 \pi \times 10^{-7} \mathrm{~Nm},+M B, -M B
Option A (375π×10−7 Nm,+MB,−MB375 \pi \times 10^{-7} \mathrm{~Nm},+M B, -M B) is correct because it correctly calculates the maximum torque and the range of potential energies using the given parameters.
279 Medium
279 of 418
Understanding +4 | -1
If the number of turns per unit length of a coil of a solenoid is doubled, the self inductance of the solenoid will
View answer and explanation
Correct option: D
Become four times
Option D (Become four times) is correct because self-inductance is proportional to the square of the number of turns per unit length.
280 Medium
280 of 418
Understanding +4 | -1
If the angular momentum of an electron is J⃗  \vec J\; then the magnitude of the magnetic moment will be
View answer and explanation
Correct option: B
eJ2m\frac{{eJ}}{{2m}}
Option B (eJ2m\frac{{eJ}}{{2m}}) is correct because the magnitude of the magnetic moment is related to the angular momentum by the formula μ=eJ2m\mu = \frac{{eJ}}{{2m}}.
281 Medium
281 of 418
Understanding +4 | -1
Assertion: The ion cannot move with speed beyond a certain limit in a cyclotron.

Reason: As velocity increases time taken by ion increases.
View answer and explanation
Correct option: C
(A) is true, (R) is false
Option C ((A) is true, (R) is false) is correct because the ion's speed is limited by the radius of the dees, and the time taken is independent of the ion's velocity.
282 Medium
282 of 418
Analyzing +4 | -1
Assertion (A): In M.C.G, the deflection 'θ\theta' is directly proportional to the strength of the current.

Reason (R): In M.C.G., the the torque experienced by the loop is BiANcos⁡θ\mathrm {BiAN}\cos\theta
View answer and explanation
Correct option: C
(A) is true, (R) is false
Option C ((A) is true, (R) is false) is correct because while the deflection is proportional to current, the torque is proportional to sin⁡θ\sin \theta, not cos⁡θ\cos \theta.
283 Medium
283 of 418
Understanding +4 | -1
This question has Statement I and Statement II. Of the four choices given after the statements, choose the one that best describes the two statements.

Statement I Higher the range, greater is the resistance of ammeter.

Statement II To increase the range of ammeter, additional shunt needs to be used across it.
View answer and explanation
Correct option: D
Statement I is false, Statement II is true
Option D (Statement I is false, Statement II is true) is correct because a higher range ammeter has lower resistance, and a shunt is used to increase the range.
284 Medium
284 of 418
Understanding +4 | -1
The force experienced by a charge qq having velocity vv parallel to a magnetic field BB which acts on the charge
View answer and explanation
Correct option: B
zero
Option B (zero) is correct because the magnetic force on a charge moving parallel to a magnetic field is zero, as the angle between the velocity and magnetic field is 0 degrees.
285 Medium
285 of 418
Analyzing +4 | -1
Statement-1: Two long parallel conductors carrying currents in the same direction experience a force of attraction.
Statement-2: The magnetic produced in the space between the conductors are in the same direction.
View answer and explanation
Correct option: C
Statement- 1 is true, Statement-2 is false
Option C is correct because Statement-1 is true, and Statement-2 is false. Parallel currents attract, and the magnetic fields between them are in opposite directions.
286 Medium
286 of 418
Remembering +4 | -1
The current that must flow through the coil of a galvanometer so as to produce a deflection of one division on its scale is called
View answer and explanation
Correct option: B
Current sensitivity of the galvanometer
Option B (Current sensitivity of the galvanometer) is correct because current sensitivity is defined as the deflection produced per unit current flowing through the galvanometer.
287 Medium
287 of 418
Understanding +4 | -1
Assertion: A current carrying toroid is freely suspended in uniform magnetic field then torque acts on it is always zero.
Reason: The magnetic moment of current carrying toroid is zero.
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 a current carrying toroid has zero magnetic moment, resulting in zero torque in a uniform magnetic field.
288 Medium
288 of 418
Analyzing +4 | -1
A current carrying straight wire is kept along
the axis of a circular loop carrying a current.
The straight wire
View answer and explanation
Correct option: D
Will not exert any force on the circular loop
Option D (Will not exert any force on the circular loop) is correct because the magnetic field due to the straight wire is along the circumference of the loop, resulting in zero force on the loop.
289 Medium
289 of 418
Application +4 | -1
A long solenoid of length 40 cm40 \mathrm{~cm} having 240 turns carries a current of 5 A5 \mathrm{~A}. The magnetic field at the centre of the solenoid is
View answer and explanation
Correct option: C
3.77×10−3 T3.77 \times 10^{-3} \mathrm{~T}
Option C (3.77×10−3 T3.77 \times 10^{-3} \mathrm{~T}) is correct because it is calculated using the formula for the magnetic field at the center of a long solenoid.
290 Medium
290 of 418
Understanding +4 | -1
If a velocity has both perpendicular and
parallel components while moving through a
magnetic field, what is the path followed by a
charged particle ?
View answer and explanation
Correct option: D
Helical
The correct answer is D because when a charged particle has both perpendicular and parallel velocity components in a magnetic field, it follows a helical path.
291 Medium
291 of 418
Application +4 | -1
A battery of emf 12 V12 \mathrm{~V} and internal resistance 1.2Ω1.2 \Omega supplies a current through a coil of resistance 48Ω48 \Omega. A voltmeter of resistance 72Ω72 \Omega is used to measure the potential difference across the coil. What would be the reading on the voltmeter ?
View answer and explanation
Correct option: A
11.52 V
Option A (11.52 V) is correct because it is the voltage reading calculated by considering the parallel combination of the coil and voltmeter resistances and the internal resistance of the battery.
292 Medium
292 of 418
Remembering +4 | -1
A moving coil galvanometer has NN number of turns in a coil of effective area AA, it carries a current II. The magnetic field BB is radial. The torque acting on the coil is
View answer and explanation
Correct option: D
NABIN A B I
Option D (NABIN A B I) is correct because the torque on a current-carrying coil in a magnetic field is given by the formula τ=NABI\tau = N A B I.
293 Medium
293 of 418
Analyzing +4 | -1
Assertion (A): If a proton and an α\alpha-particle enter a uniform magnetic field perpendicularly with the same speed, then the time period of revolution of the α\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 because both the assertion and the reason are true, but the reason does not directly explain the assertion. The time period depends on mass and charge, and the alpha particle has twice the charge and four times the mass of a proton.
294 Medium
294 of 418
Understanding +4 | -1
Which of the following statements is wrong
View answer and explanation
Correct option: C
Ammeter is placed in parallel across the conductor in a circuit
Option C (Ammeter is placed in parallel across the conductor in a circuit) is wrong because an ammeter is always connected in series to measure current.
295 Medium
295 of 418
Understanding +4 | -1
A long hollow tube is carrying an electric current along its length distributed uniformly over its surface. The magnetic field
View answer and explanation
Correct option: C
Is zero at the axis
Option C (Is zero at the axis) is correct because, according to Ampere's law, the magnetic field inside a hollow tube carrying a uniform current is zero.
296 Medium
296 of 418
Analyzing +4 | -1
Assertion: A moving charged particle gets energy from magnetic field.
Reason: Magnetic force works on moving charged particle.
View answer and explanation
Correct option: D
If both Assertion & Reason are false
Option D is correct because a magnetic field does no work on a moving charged particle, hence it cannot give energy to it. The magnetic force is always perpendicular to the velocity.
297 Medium
297 of 418
Application +4 | -1
An electron is revolving around a proton in a circular path of diameter 0.1 nm0.1 {~nm}. It produces a magnetic field 14 T14 {~T} at a proton. Then the angular speed of the electron is
View answer and explanation
Correct option: B
4.4×1016rads−14.4 \times 10^{16} {rad} {s}^{-1}
Option B (4.4×1016rads−14.4 \times 10^{16} {rad} {s}^{-1}) is correct because it is the angular speed calculated using the magnetic field produced by the electron revolving around the proton.
298 Medium
298 of 418
Application +4 | -1
In the hydrogen atom, the electron is making 6.6×1015rps6.6 \times 10^{15} \mathrm{rps}. If the radius of orbits is 0.53×10−0.53 \times 10^{-} 10 m{ }^{10} \mathrm{~m}, then magnetic field produced at the centre of the orbit is
View answer and explanation
Correct option: B
12.5 T12.5 \mathrm{~T}
Option B (12.5 T) is correct because it is calculated using the formula for the magnetic field at the center of a circular current loop, B=μ0I2rB = \frac{\mu_0 I}{2r}, where I=qfI = qf.
299 Medium
299 of 418
Application +4 | -1
A thin magnetic needle is bent to form a semicircle. What will be the percentage change in equivalent dipole moment?
View answer and explanation
Correct option: D
(2−ππ)×100\left(\frac{2-\pi}{\pi}\right) \times 100
Option D ((2−ππ)×100\left(\frac{2-\pi}{\pi}\right) \times 100) is correct because it represents the percentage change in the magnetic dipole moment after bending the needle.
300 Easy
300 of 418
Remembering +4 | -1
The magnetic moment of an electron (e) revolving in an orbit around nucleus with an orbital angular momentum is given by:
View answer and explanation
Correct option: B
μL→=−eL→2m\overrightarrow{\mu_L}=-\frac{\overrightarrow{e L}}{2 m}
Option B (μL→=−eL→2m\overrightarrow{\mu_L}=-\frac{\overrightarrow{{e L}}}{{2 m}}) is correct because the magnetic moment of an electron is related to its angular momentum by this equation, with the negative sign indicating the anti-parallel direction.
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