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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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Showing 241-270 of 385 questions

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241 Medium
241 of 385
Understanding +4 | -1
A charged particle moves in a circular path in a uniform magnetic field. If its speed is reduced, then its time period will
View answer and explanation
Correct option: C
remain same
Option C (remain same) is correct because the time period of a charged particle in a uniform magnetic field is independent of its speed.
242 Medium
242 of 385
Understanding +4 | -1
The materials for the construction of electro magnets should have
View answer and explanation
Correct option: A
High initial permeability
The correct answer is A (High initial permeability) because electromagnets need to magnetize quickly and strongly, which is facilitated by high initial permeability.
243 Medium
243 of 385
Application +4 | -1
An electron moving with velocity 2×107 m/s2 \times 10^7 \mathrm{~m} / \mathrm{s}, describes a circle in a magnetic field of strength 2×10−2T.2 \times {10^{ - 2}}T. If em\frac{\mathrm{e}}{\mathrm{m}} of electron is 1.76×1011C/kg1.76 \times 10^{11} \mathrm{C} / \mathrm{kg}.
Then the diameter of the circle will be
View answer and explanation
Correct option: B
1.1 cm1.1 \mathrm{~cm}
Option B (1.1 cm1.1 \mathrm{~cm}) is correct because the diameter of the circular path is calculated using the given velocity, magnetic field, and charge-to-mass ratio.
244 Medium
244 of 385
Application +4 | -1
A solenoid of length 22.5 cm has a total of 900 turns and carries a current of 0.8 A. What is the magnetizing field H near the centre and far away from the end of the solenoid?
View answer and explanation
Correct option: A
3200  Am−13200\,\,A{m^{ - 1}}
Option A (3200,Am−13200 \\,A{m^{ - 1}}) is correct because it is calculated directly from the formula for the magnetizing field H inside a solenoid.
245 Medium
245 of 385
Application +4 | -1
The magnetic field at the centre of a circular coil of radius rr, due to current II flowing through it, is BB. The magnetic field at a point along the axis at a distance r2\frac{r}{2} from the centre is:
View answer and explanation
Correct option: C
(25)3B{\left( {\frac{2}{{\sqrt 5 }}} \right)^3}B
Option C ((25)3B{\left( {\frac{2}{{\sqrt 5 }}} \right)^3}B) is correct because it correctly applies the formula for the magnetic field on the axis of a circular coil.
246 Medium
246 of 385
Understanding +4 | -1
Statement-1 : The sensitivity of a moving coil galvanometer is increased by placing a suitable magnetic material as a core inside the coil.

and
Statement-2 : Soft iron has a high magnetic permeability and cannot be easily magnetized or demagnetized.
View answer and explanation
Correct option: C
Statement - 1 is True, Statement-2 is False
Statement - 1 is True, Statement-2 is False. Soft iron is easily magnetized and demagnetized, making statement 2 incorrect.
247 Medium
247 of 385
Analyzing +4 | -1
A positively charged particle moves in a region having a uniform magnetic field and uniform electric field in same direction. At time t=0t=0, the velocity of the particle is perpendicular to the field direction. The subsequent path of the particle will be
View answer and explanation
Correct option: D
A helix with increasing pitch
The correct answer is D because the magnetic field causes a circular path, and the electric field accelerates the particle along that path, resulting in a helix with increasing pitch.
248 Medium
248 of 385
Application +4 | -1
A solenoid having 2000 turns per meter has a core of area 4 cm24 \mathrm{~cm}^2 and relative permeability 220. It carries a current of 5 A5 \mathrm{~A}. The which statement among the following is true?
View answer and explanation
Correct option: A
Magnetic intensity is 104 A.m−110^4 \mathrm{~A} . \mathrm{m}^{-1}. Magnetic field is 88π×10−2 T88 \pi \times 10^{-2} \mathrm{~T} and Pole strength is 876 A.m876 \mathrm{~A} . \mathrm{m}
Option A (Magnetic intensity is 104 A.m−110^4 \mathrm{~A} . \mathrm{m}^{-1}. Magnetic field is 88π×10−2 T88 \pi \times 10^{-2} \mathrm{~T} and Pole strength is 876 A.m876 \mathrm{~A} . \mathrm{m}) is correct because it is calculated using the formulas for magnetic intensity, magnetic field, and pole strength in a solenoid with a core.
249 Medium
249 of 385
Application +4 | -1
A 0.1 m long conductor carrying a current of 50A is perpendicular to a magnetic field of 1.25 mT. The mechanical power to move the conductor with a speed of 1m/s is
View answer and explanation
Correct option: B
6.25 mw
Option B (6.25 mw) is correct because it is calculated by finding the force on the conductor due to the magnetic field and then multiplying it by the velocity to find the power.
250 Medium
250 of 385
Understanding +4 | -1
Assertion: A current element is a source of magnetic field.
Reason: A moving charge is a source of magnetic field.
View answer and explanation
Correct option: A
If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion
Option A (If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion) is correct because moving charges (current elements) create magnetic fields.
251 Medium
251 of 385
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.
252 Medium
252 of 385
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.
253 Medium
253 of 385
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.
254 Medium
254 of 385
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.
255 Medium
255 of 385
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.
256 Medium
256 of 385
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.
257 Medium
257 of 385
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.
258 Medium
258 of 385
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}}.
259 Medium
259 of 385
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.
260 Medium
260 of 385
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.
261 Medium
261 of 385
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.
262 Medium
262 of 385
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.
263 Medium
263 of 385
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.
264 Medium
264 of 385
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.
265 Medium
265 of 385
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.
266 Medium
266 of 385
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.
267 Medium
267 of 385
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.
268 Medium
268 of 385
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.
269 Medium
269 of 385
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.
270 Medium
270 of 385
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.
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