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JEE • PHYSICS • MAGNETIC EFFECT OF CURRENT

Magnetic effect of current practice · Medium

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271 Medium
271 of 385
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.
272 Medium
272 of 385
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.
273 Medium
273 of 385
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.
274 Medium
274 of 385
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.
275 Medium
275 of 385
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.
276 Medium
276 of 385
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.
277 Medium
277 of 385
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.
278 Medium
278 of 385
Understanding +4 | -1
Ions of different momentum (pp), having the same charge, enter normally a uniform magnetic field. The radius of the orbit of an ion is proportional to
View answer and explanation
Correct option: A
PP
Option A (PP) is correct because the radius of the orbit of an ion in a magnetic field is directly proportional to its momentum.
279 Medium
279 of 385
Understanding +4 | -1
With the help of potentiometer, we can determine the value of emf of a given cell. The sensitivity of the potentiometer is

(A) Directly proportional to the length of the potentiometer wire

(B) Directly proportional to the potential gradient of the wire

(C) inversely proportional to the potential gradient of the wire

(D) inversely proportional to the length of the potentiometer wire

Choose the correct option for the above statements:
View answer and explanation
Correct option: C
A and C only
Option A and C only is correct because the sensitivity of a potentiometer is directly proportional to the length of the potentiometer wire and inversely proportional to the potential gradient of the wire.
280 Medium
280 of 385
Application +4 | -1
A magnetic field of (4.0×10−3k⃗)T\left(4.0 \times 10^{-3} \vec{k}\right) \mathrm{T} exerts a force of (4.0i⃗+3.0j⃗)×10−10 N(4.0 \vec{i}+3.0 \vec{j}) \times 10^{-10} \mathrm{~N} on a particle having a charge of 1.0×10−9C1.0 \times 10^{-9} \mathrm{C} and going in the x−yx-y plane. Find the velocity of the particle.
View answer and explanation
Correct option: C
(−75i^+100j^)m/s(-75 \hat{i}+100 \hat{j}) \mathrm{m/s}
Option C ((-75 \hat{i}+100 \hat{j}) m/s) is correct because it is the calculated velocity of the particle based on the given magnetic field and force.
281 Medium
281 of 385
Analyzing +4 | -1
Assertion: Resistance of an ammeter is less than the resistance of a milli ammeter.

Reason: Value of shunt required in case of ammeter is more than a milli ammeter.
View answer and explanation
Correct option: B
If both Assertion and Reason are true but Reason is not correct explanation of Assertion.
Option B is correct. Both the assertion and reason are true, but the reason does not directly explain why an ammeter's resistance is less than a milliammeter's.
282 Medium
282 of 385
Understanding +4 | -1
Two parallel wires carrying currents in
opposite directions
View answer and explanation
Correct option: A
repel each other
Option A is correct because parallel wires carrying currents in opposite directions create magnetic fields that result in a repulsive force between the wires.
283 Medium
283 of 385
Understanding +4 | -1
A magnetic field exerts no force on
View answer and explanation
Correct option: D
stationary charge
Option D (stationary charge) is correct because a magnetic field only exerts a force on moving charges.
284 Medium
284 of 385
Understanding +4 | -1
Assertion: Cyclotron is a accelerating device.
Reason: Neutrons can be accelerated by the cyclotron.
View answer and explanation
Correct option: C
If Assertion is True but the Reason is False
Option C (If Assertion is True but the Reason is False) is correct. Cyclotrons accelerate charged particles, but neutrons, being neutral, cannot be accelerated by them.
285 Medium
285 of 385
Understanding +4 | -1
A charged particle moving in a magnetic field
experiences a resultant force
View answer and explanation
Correct option: C
in the direction perpendicular to both the field
and its velocity
Option C (in the direction perpendicular to both the field and its velocity) is correct because the magnetic force is given by the cross product of velocity and magnetic field.
286 Medium
286 of 385
Application +4 | -1
A cyclotron is used to accelerate protons. If the operating magnetic field is 1.0 T1.0 \mathrm{~T} and the radius of the cyclotron 'dees' is 60 cm60 \mathrm{~cm}, the kinetic energy of the accelerated protons in MeV\mathrm{MeV} will be

[\left[\right. Use mp=1.6×10−27 kg,e=1.6×10−19Cm_p=1.6 \times 10^{-27} \mathrm{~kg}, e=1.6 \times 10^{-19} \mathrm{C} ]
View answer and explanation
Correct option: B
18
Option B (18) is the correct answer because it is calculated using the formula for kinetic energy in a cyclotron, considering the magnetic field and radius.
287 Medium
287 of 385
Application +4 | -1
A proton, a deuteron and α\alpha particle accelerated through the same potential difference enter the region of uniform magnetic field, moving at right angles to magnetic field. The ratio of their kinetic energies will be
View answer and explanation
Correct option: B
1:1:21: 1: 2
Option B (1:1:21: 1: 2) is correct because the kinetic energy gained by a charged particle accelerated through a potential difference is proportional to its charge.
288 Medium
288 of 385
Application +4 | -1
A positive charged particle qq, mass mm is dropped from certain point above ground. A constant horizontal magnetic field of magnitude BB is present. If charge falls down by height hh, then speed of charge is (neglect effect of air)
View answer and explanation
Correct option: A
2gh\sqrt{2 g h}
Option A (2gh\sqrt{2 g h}) is correct because the speed of the charged particle depends only on the height fallen due to gravity, as the magnetic force does no work.
289 Medium
289 of 385
Analyzing +4 | -1
In a region an electron is moving in a circle. What can be the possible field there?
View answer and explanation
Correct option: D
Both (2) and (3)
Option D (Both (2) and (3)) is correct because an electron can move in a circle in a time-varying electric field (due to induced magnetic field) or a uniform magnetic field.
290 Medium
290 of 385
Application +4 | -1
The maximum current that can be measured by a galvanometer of resistance 40  Ω\mathrm{40}\,\,\mathrm{\Omega } is 10  mA\mathrm{10}\,\,\mathrm{mA}. It is converted into a voltmeter that can read upto 50  V\mathrm{50}\,\,\mathrm{V}. The resistance to be connected is series with the galvanometer (in ohm) is
View answer and explanation
Correct option: D
4960
Option D (4960) is correct because the series resistance required to convert the galvanometer into a voltmeter is calculated using the formula R = V/Ig - G.
291 Medium
291 of 385
Understanding +4 | -1
Two galvanometers 'A' and 'B' require currents of 4 mA4 \mathrm{~mA} and 7 mA7 \mathrm{~mA}, respectively to produce the same deflection of 20 divisions. If ' SA\mathrm{S}_{\mathrm{A}} ' and ' SB\mathrm{S}_{\mathrm{B}} ' are their sensitivities, respectively, then
View answer and explanation
Correct option: D
SA>SB\mathrm{S}_{\mathrm{A}} > \mathrm{S}_{\mathrm{B}}
Option D (SA>SB\mathrm{S}_{\mathrm{A}} > \mathrm{S}_{\mathrm{B}}) is correct because sensitivity is inversely proportional to the current required for the same deflection, and galvanometer A requires less current.
292 Medium
292 of 385
Analyzing +4 | -1
Assertion: Cyclotron is used to accelerate heavy charged particles, such as proton. It not suitable for accelerating light charged particles, such as electrons.

Reason: Specific charge of proton is more as compared to electron.
View answer and explanation
Correct option: C
Assertion is true & Reason is false
Option C (Assertion is true & Reason is false) is correct because while the cyclotron is indeed unsuitable for accelerating light charged particles like electrons, the reason is due to relativistic mass increase, not specific charge.
293 Medium
293 of 385
Application +4 | -1
Two wires are held perpendicular to the plane of paper and are 5 m5 \mathrm{~m} apart. They carry currents of 2.5 A2.5 \mathrm{~A} and 5 A5 \mathrm{~A} in same direction. Then, the magnetic field strength (B)(B) at a point midway between the wires will be
View answer and explanation
Correct option: B
μ02πT\frac{\mu_0}{2 \pi} \mathrm{T}
Option B (μ02πT\frac{{\mu_0}}{{2 \pi}} \mathrm{{T}}) is correct because the magnetic field is the difference between the fields created by each wire at the midpoint.
294 Medium
294 of 385
Analyzing +4 | -1
Assertion: A circular loop carrying current lies in XY plane with its centre at origin having a magnetic flux in negative Z-axis.

Reason: Magnetic flux direction is independent of the direction of current in the conductor.
View answer and explanation
Correct option: C
(A) is true, (R) is false
The correct answer is A because the electrostatic force between the plates of an isolated capacitor depends only on the charge and area, not the distance between the plates.
295 Medium
295 of 385
Application +4 | -1
A particle moving in a circle of radius RR with a uniform speed takes a time TT to complete one revolution. If this particle were projected with the same speed at an angle 'θ\theta' to the horizontal, the maximum height attained by it equals 4R.4R. The angle of projection 'θ\theta' is then given by:
View answer and explanation
Correct option: B
θ=sin⁡−1(2gT2π2R)12\mathrm{\theta } = {\sin ^{ - 1}}{\left( {\frac{{2g{T^2}}}{{{\pi ^2}R}}} \right)^{\frac{1}{2}}}
Option B (θ=sin⁡−1(2gT2π2R)12\mathrm{\theta } = {\sin ^{ - 1}}{\left( {\frac{{2g{T^2}}}{{{\pi ^2}R}}} \right)^{\frac{1}{2}}}) is correct because it correctly relates the angle of projection to the time period, radius, and acceleration due to gravity.
296 Medium
296 of 385
Understanding +4 | -1
Sensitivity of potentiometer can be increased by
View answer and explanation
Correct option: B
Increasing the length of the potentiometer wire
Option B is correct because increasing the length of the potentiometer wire decreases the potential gradient, which increases the sensitivity of the potentiometer.
297 Medium
297 of 385
Analyzing +4 | -1
Assertion: If the current in a solenoid is
reversed in direction while keeping the same
magnitude, the magnetic field energy stored in the
solenoid decreases.
Reason: Magnetic field energy density is
proportional to square of current
View answer and explanation
Correct option: D
If both the Assertion and Reason are
incorrect.
Option D (If both the Assertion and Reason are incorrect) is correct because reversing the current direction does not decrease the magnetic field energy, and the energy density is proportional to the square of the magnetic field, not necessarily the current directly.
298 Medium
298 of 385
Application +4 | -1
A straight conductor carries a current of 5 A5 \mathrm{~A}. An electron travelling with a speed of 5×1065 \times 10^6 ms−1\mathrm{ms}^{-1} parallel to the wire at a distance of 0.1 m0.1 \mathrm{~m} from the conductor, experiences a force of
View answer and explanation
Correct option: C
8×10−18 N8 \times 10^{-18} \mathrm{~N}
Option C (8×10−18 N8 \times 10^{-18} \mathrm{~N}) is correct because the force on the electron is calculated using the magnetic field due to the wire and the Lorentz force formula.
299 Medium
299 of 385
Application +4 | -1
The magnetic field applied on a cyclotron is 3.5 T3.5 {~T}. What will be the frequency of electric field that must be applied between the Dee's in order to accelerate a proton?
View answer and explanation
Correct option: C
5.57×107 Hz5.57 \times 10^7 {~Hz}
Option C (5.57×107 Hz5.57 \times 10^7 {~Hz}) is correct because the frequency of the electric field is calculated using the magnetic field and the charge-to-mass ratio of the proton.
300 Medium
300 of 385
Application +4 | -1
A proton is moving in a uniform magnetic field BB in a circular path of radius a in a direction perpendicular to z-axis along which field BB exists. Calculate the angular momentum. If the radius is a charge on proton is ee.
View answer and explanation
Correct option: C
a2eB\mathrm{a}^2 \mathrm{eB}
Option C (a2eB\mathrm{a}^2 \mathrm{eB}) is correct because it is the angular momentum calculated using the radius, charge, and magnetic field.
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