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JEE • PHYSICS • MAGNETIC EFFECT OF CURRENT

Magnetic effect of current practice · Medium

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These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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301 Medium
301 of 385
Analyzing +4 | -1
As shown in figure a square loop made from a uniform wire. Calculate the magnetic field at the centre of the square if a battery source is connected between the points 'A' & 'C'.
View answer and explanation
Correct option: A
Zero
Option A (Zero) is correct because the current divides equally at point A, and the magnetic fields due to opposite sides of the square cancel out at the center.
302 Medium
302 of 385
Application +4 | -1
A charged particle is moving in a uniform magnetic field penetrates a layer of lead and thereby loss half of its kinetic energy. Then the radius of curvature of its path is.
View answer and explanation
Correct option: C
Reduced to 12\frac{1}{\sqrt{2}} times of its initial values
Option C (Reduced to 12\frac{1}{\sqrt{2}} times of its initial values) is correct because the radius of curvature is proportional to the square root of the kinetic energy.
303 Medium
303 of 385
Analyzing +4 | -1
Assertion: The trajectory of neutron when it is projected perpendicular to an electric field is parabola.
Reason: A moving charge entering parallel to the magnetic field lines moves in a circular path.
View answer and explanation
Correct option: D
If both Assertion & Reason are false
Option D (If both Assertion & Reason are false) is correct because a neutron projected perpendicular to an electric field follows a parabolic path, and a moving charge entering parallel to a magnetic field experiences no force, thus moves in a straight line.
304 Medium
304 of 385
Analyzing +4 | -1
A cubical region of space is filled with some uniform electric and magnetic fields. An electron enters, the cube across one of its faces with velocity v{v} and a positron enters via opposite face with velocity −v-{v}. At this instant,
View answer and explanation
Correct option: C
Both particles gain or loose energy at the same rate
Option C is correct because both particles have the same magnitude of charge and are moving with the same speed in opposite directions, so the rate at which they gain or lose energy due to the magnetic field is the same.
305 Medium
305 of 385
Understanding +4 | -1
A conducting circular loop of radius rr carries a constant current ii. It is placed in uniform magnetic field B⃗\vec{B}, such that B⃗\vec{B} is perpendicular to plane of the loop. Find the magnetic force acting on the loop.
View answer and explanation
Correct option: C
Zero
The correct answer is C (Zero) because the net magnetic force on a closed current-carrying loop in a uniform magnetic field is always zero.
306 Medium
306 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 resistance to be connected in series with the galvanometer is calculated using the formula for converting a galvanometer into a voltmeter.
307 Medium
307 of 385
Application +4 | -1
A proton projected in a magnetic field of 0.020 T travels along a helical path of radius 5.0 cm and pitch 20 cm. Find the components of the velocity of the proton along and perpendicular to the magnetic field. Take the mass of the proton =1.6×10−27 kg=1.6 \times 10^{-27} \mathrm{~kg}.
View answer and explanation
Correct option: A
1×105 m/s;  6.4×104 m/s1 \times {10^5}\,m/s;\,\,6.4 \times {10^4}\,m/s
Option A (1×105 m/s;  6.4×104 m/s1 \times {10^5}\,m/s;\,\,6.4 \times {10^4}\,m/s) is correct because the perpendicular component is found using the radius of the helix, and the parallel component is found using the pitch.
308 Medium
308 of 385
Application +4 | -1
An ammeter reads upto 1 A. Its internal resistance is 0.81  Ω0.81\,\,\Omega. To increase the range to 10 A, the value of the required shunt is-
View answer and explanation
Correct option: D
0.09  Ω0.09\,\,\Omega
Option D (0.09,Ω0.09\\,\Omega) is correct because it is calculated using the shunt resistance formula for extending ammeter range.
309 Medium
309 of 385
Understanding +4 | -1
If a charged particle is projected with a velocity ' vv ' in a magnetic field ' BB ', then power developed by the field is
View answer and explanation
Correct option: C
zero
Option C (zero) is correct because the magnetic force on a moving charge is always perpendicular to the velocity, so the work done and thus the power developed by the magnetic field is zero.
310 Medium
310 of 385
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 in the magnetic meridian, not at right angles.
311 Medium
311 of 385
Understanding +4 | -1
In a current carrying long solenoid, the field produced does not depend upon
View answer and explanation
Correct option: C
Radius of solenoid
Radius of solenoid is correct because the magnetic field inside a long solenoid depends on the number of turns per unit length and the current, but not the radius.
312 Medium
312 of 385
Understanding +4 | -1
An electron flies into a homogeneous magnetic field of 10−3 T10^{-3} \mathrm{~T} perpendicular to the force lines. The velocity of the electron is v=4×107 m s−1{v}=4 \times 10^7 \mathrm{~m} \mathrm{~s}^{-1}. What is the tangential acceleration of electron in the magnetic field?
View answer and explanation
Correct option: D
Zero
Option D is correct because the magnetic force on a moving charge does not change its speed, only its direction, resulting in zero tangential acceleration.
313 Medium
313 of 385
Understanding +4 | -1
Given below are two statements: One is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A): In an uniform magnetic field, speed and energy remains the same for a moving charged particle.
Reason (R): Moving charged particle experiences magnetic force perpendicular to its direction of motion.
View answer and explanation
Correct option: A
Both (A) and (R) are true and (R) is the correct explanation of (A).
Both Assertion and Reason are true, and Reason correctly explains Assertion because the magnetic force being perpendicular to the velocity does no work, thus keeping speed and energy constant.
314 Medium
314 of 385
Understanding +4 | -1
An electron with mass mm, velocity vv and charge ee describing half a revolution in a circle of radius rr in a magnetic field BB, will experience change in energy equal to
View answer and explanation
Correct option: D
Zero
Option D (Zero) is correct because the magnetic force is always perpendicular to the velocity, resulting in no work done and no change in energy.
315 Medium
315 of 385
Analyzing +4 | -1
Assertion (A) : A charged particle is moving in a circle with constant speed in uniform magnetic field. If we increase the speed of particle to twice, its acceleration will become four times.

Reason (R) : In circular path of radius RR with constant speed vv, acceleration is given by v2R\frac{v^2}{R}.
View answer and explanation
Correct option: D
If Assertion is false but the Reason is true.
Option D is correct because the assertion is false (acceleration becomes twice, not four times) and the reason is true (acceleration is given by v²/R).
316 Medium
316 of 385
Understanding +4 | -1
Three moving coil galvanometers
A,B  and  CA,\mathrm{ }B\,\,and\,\,C
are made of coils of three different material having torsional constant 1.8  ×  10−8,  2.8  ×  10−81.8\,\, \times \,\,{10^{ - 8}},\,\,2.8\,\, \times \,\,{10^{ - 8}}and 3.8  ×  10−83.8\,\, \times \,\,{10^{ - 8}} respectively. If the three galvanometers are identical in all other respect, then in which of the above cases sensitivity is maximum?
View answer and explanation
Correct option: A
A\begin{array}{llllllllllllllllllll} A \end{array}
Option A is correct because the current sensitivity is inversely proportional to the torsional constant, and galvanometer A has the lowest torsional constant.
317 Medium
317 of 385
Understanding +4 | -1
The coil of a moving coil galvanometer is wound over a metal frame in order to
View answer and explanation
Correct option: B
Provide electromagnetic damping
Option B (Provide electromagnetic damping) is correct because winding the coil over a metal frame in a moving coil galvanometer induces eddy currents, which provide electromagnetic damping.
318 Medium
318 of 385
Application +4 | -1
A wire of length LL meter carrying a current of II ampere is bent in the form of a circle. Its magnetic moment is:
View answer and explanation
Correct option: D
IL24π A−m2\frac{{IL}^2}{4 \pi} \mathrm{~A}-\mathrm{m}^2
Option D (IL24π A−m2\frac{{IL}^2}{4 \pi} \mathrm{~A}-\mathrm{m}^2) is correct because it is derived from calculating the area of the circle formed by the wire and then multiplying by the current.
319 Medium
319 of 385
Understanding +4 | -1
A positively charged particle enters in a uniform magnetic field with uniform velocity vv. If angle between velocity and magnetic field is 180∘,180^{\circ}, the trajectory will be
View answer and explanation
Correct option: A
Straight line
Option A (Straight line) is correct because when the angle between velocity and magnetic field is 180 degrees, the magnetic force is zero, resulting in no deflection.
320 Medium
320 of 385
Understanding +4 | -1
A metal rod is moved with a constant velocity vv in a magnetic field. A potential difference appears across the two ends:
View answer and explanation
Correct option: D
none of these
Option D (none of these) is correct because a potential difference appears only when the velocity, length, and magnetic field are mutually perpendicular.
321 Medium
321 of 385
Understanding +4 | -1
Assertion (A): Out of galvanometer, ammeter and voltmeter, resistance of ammeter is lowest and resistance of voltmeter is highest.

Reason (R): An ammeter is connected in series and a voltmeter is connected in parallel, in a circuit.
View answer and explanation
Correct option: B
(A) and (R) are true and (R) is not correct explanation of (A)
Option B ((A) and (R) are true and (R) is not correct explanation of (A)) is correct because both the assertion and the reason are true, but the reason doesn't fully explain the assertion.
322 Medium
322 of 385
Application +4 | -1
A particle of charge qq and mass mm starts moving from the origin under the action of an electric field, E=E0i^E=E_0 \hat{{i}} and B=B0i^B=B_0 \hat{{i}} with a velocity, v=v0j^v=v_0 \hat{{j}}. The speed of the particle will become 52v0\frac{\sqrt{5}}{2} v_0 after a time
View answer and explanation
Correct option: B
mv02qE\frac{m v_0}{2 q E}
Option B (mv02qE\frac{m v_0}{2 q E}) is correct because it is derived from calculating the time when the particle's speed reaches the given value under the influence of electric and magnetic fields.
323 Medium
323 of 385
Application +4 | -1
A galvanometer has a resistance of 30 Ω30\,\Omega and current of 2 mA2 \mathrm{~mA} is needed to give full scale deflection. What is the resistance needed and how is it to be connected to convert the galvanometer into voltmeter of 0.2 volt range.
View answer and explanation
Correct option: A
70 Ω70\,\Omega
Option A (70 Ω70\,\Omega) is correct because it is the resistance needed to be connected in series with the galvanometer to convert it into a voltmeter of 0.2 volt range.
324 Medium
324 of 385
Application +4 | -1
A length of wire carries a steady current ii. It is bent first to form a circular plane coil of one turn. The same length is now bent more sharply to give three loops of smaller radius. The magnetic field at the centre caused by the same current is
View answer and explanation
Correct option: D
Nine times of its initial value
Option D (Nine times of its initial value) is correct because the magnetic field at the center of a coil is proportional to the square of the number of turns when the length of the wire is constant.
325 Medium
325 of 385
Understanding +4 | -1
An electron is projected along the axis of
circular conductor carrying current I. Electron
will experience
View answer and explanation
Correct option: A
no force.
Option A (no force.) is correct because the force on a moving charge in a magnetic field is given by F=qvBsin⁡θF = qvB\sin\theta, and since the electron is moving along the axis, the angle θ\theta between the velocity and the magnetic field is 0, making the force zero.
326 Medium
326 of 385
Application +4 | -1
Calculate the torque on a 100 turn rectangular coil of length 40 cm40 \mathrm{~cm} and breadth 20 cm20 \mathrm{~cm}, carrying a current of 10 A10 \mathrm{~A}, when placed making an angle of 60∘60^{\circ} with a magnetic field of 3T3 T.
View answer and explanation
Correct option: A
120 Nm
Option A (120 Nm) is correct because it is the calculated torque on the coil using the formula τ=NIBAsin⁡θ\tau = NIBA \sin \theta.
327 Medium
327 of 385
Application +4 | -1
A charged particle with specific charge ss moves undeflected through a region of space containing mutually perpendicular and uniform electric and magnetic fields EE and BB. When the electric field is switched off, the particle will move in a circular path of radius
View answer and explanation
Correct option: D
EB2s\frac{E}{B^2 s}
Option D (EB2s\frac{E}{B^2 s}) is correct because it is derived from equating the electric and magnetic forces and then using the formula for the radius of a circular path in a magnetic field.
328 Medium
328 of 385
Understanding +4 | -1
The momentum of a charged particle moving in a perpendicular magnetic field depends on
View answer and explanation
Correct option: D
All of the above
Option D (All of the above) is correct because the momentum of a charged particle in a magnetic field depends on its charge, the field strength, and the radius of its path.
329 Medium
329 of 385
Application +4 | -1
Two long Straight wires each carrying an electric Current of 5 A5 \mathrm{~A} are kept parallel to each other at a separation of 2.5 cm2.5 \mathrm{~cm}. Find the magnitude of the magnetic force experiment by 10 cm10 \mathrm{~cm} of a wire.
View answer and explanation
Correct option: B
3.5×10−6 N3.5 \times 10^{-6} \mathrm{~N}
The correct answer is C (2.0×10−5 N2.0 \times 10^{-5} \mathrm{~N}) because it is calculated using the formula for the magnetic force between two parallel current-carrying wires.
330 Medium
330 of 385
Application +4 | -1
A long straight wire with a circular cross-section having radius RR, is carrying a steady current II. The current II is uniformly distributed across this cross-section. Then the variation of magnetic field due to current II with distance r (r<R)r\,(r < R) from its centre will be
View answer and explanation
Correct option: B
B∝rB \propto r
Option B (B∝rB \propto r) is correct because, according to Ampere's law, the magnetic field inside a long straight wire with uniform current distribution is directly proportional to the distance from the center.
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