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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 1-30 of 385 questions
1 Medium
1 of 385
Understanding +4 | -1
A needle made of bismuth is suspended freely in a magnetic field. The angle which the needle makes with the magnetic field is:
View answer and explanation
Correct option: C
90∘90^{\circ}
Option C (90∘90^{\circ}) is correct because bismuth is a diamagnetic substance, and diamagnetic materials align themselves perpendicular to an external magnetic field.
2 Medium
2 of 385
Analyzing +4 | -1
Assertion: If a charged particles is moving in perpendicular uniform magnetic field, its kinetic energy does not change
Reason: Velocity of charge particle is not changing in magnetic field.
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 because the kinetic energy remains constant, but the velocity direction changes in a magnetic field.
3 Medium
3 of 385
Analyzing +4 | -1
Pick out the wrong statement :
View answer and explanation
Correct option: D
An electron moving in the direction of the
electric field gains KE
The correct answer is D because an electron moving in the direction of the electric field loses kinetic energy as it moves against the electric force.
4 Medium
4 of 385
Application +4 | -1
A steady current is flowing in a circular coil of radius RR, made up of a thin conducting wire. The magnetic field at the center of the loop is BLB_L. Now, a circular loop of radius R/nR/n is made from the same wire without changing its length, by unfolding and refolding the loop, and the same current is passed through it. If new magnetic field at the center of the coil is BCB_C, then the ratio BL/BCB_L / B_C is
View answer and explanation
Correct option: A
1:n21: n^2
Option A (1:n21: n^2) is correct because the magnetic field is inversely proportional to the radius and directly proportional to the number of turns, resulting in a ratio of 1/n21/n^2.
5 Medium
5 of 385
Application +4 | -1
Two particles XX and YY having equal charges, after being accelerated through the same potential difference, enter a region of uniform magnetic field and describe circular path of radius R1R_1 and R2R_2 respectively. The ratio of mass of X{X} to that of Y{Y} is :
View answer and explanation
Correct option: C
(R1R2)2\left(\frac{R_1}{R_2}\right)^2
Option C ((R1R2)2\left(\frac{R_1}{R_2}\right)^2) is correct because the ratio of masses is proportional to the square of the ratio of the radii when the charges and potential difference are the same.
6 Medium
6 of 385
Understanding +4 | -1
When a current carrying coil is placed in a uniform magnetic field with its magnetic moment anti-parallel to the field

a) Torque on it is maximum
b) Torque on it is zero

c) Potential energy is maximum
d) Dipole is in unstable equilibrium
View answer and explanation
Correct option: A
b, c, d are correct
Option A (b, c, d are correct) is correct because when the magnetic moment is anti-parallel to the field, the torque is zero, potential energy is maximum, and the dipole is in unstable equilibrium.
7 Medium
7 of 385
Understanding +4 | -1
If an electron is going in the direction of magnetic field B⃗\vec{\text{B}} with the velocity V⃗,\vec{\text{V}}, then the force on electron is
View answer and explanation
Correct option: A
Zero
Option A (Zero) is correct because the force on a moving charge in a magnetic field is given by F→=q(V→×B→)\overrightarrow{\mathrm{F}}=\mathrm{q}(\overrightarrow{\mathrm{V}} \times \overrightarrow{\mathrm{B}}), and when the velocity and magnetic field are in the same direction, the angle between them is 0, so the force is zero.
8 Medium
8 of 385
Application +4 | -1
The magnetic needle of a tangent galvanometer is deflected at an angle 30∘30^\circ due to a magnet. The horizontal component of earth’s magnetic field 0.34×10−40.34 \times {10^{ - 4}} T is along the plane of the coil. The magnetic intensity is
View answer and explanation
Correct option: C
1.96×10−5\times {10^{ - 5}} T
Option C (1.96×10−5\times {10^{ - 5}} T) is correct because the magnetic intensity is calculated using the tangent law.
9 Medium
9 of 385
Understanding +4 | -1
The magnetic force on a charged particle
moving in the field does not work, because
View answer and explanation
Correct option: A
kinetic energy of the charged particle does not
change
Option A (kinetic energy of the charged particle does not change) is correct because the magnetic force is always perpendicular to the velocity, so it does no work, and the kinetic energy remains constant.
10 Medium
10 of 385
Remembering +4 | -1
The magnetic field dB⃗d \vec{B} due to a small current element dl⃗d \vec{l} at distance r⃗\vec{r} and element carrying current ii is
View answer and explanation
Correct option: D
dB⃗=μ0i4π(dl⃗×r⃗r3)d \vec{B}=\frac{\mu_0 i}{4 \pi}\left(\frac{d \vec{l} \times \vec{r}}{r^3}\right)
Option D (dB⃗=μ0i4π(dl⃗×r⃗r3)d \vec{B}=\frac{\mu_0 i}{4 \pi}\left(\frac{d \vec{l} \times \vec{r}}{r^3}\right)) is correct because it represents the Biot-Savart law in vector form, describing the magnetic field due to a small current element.
11 Medium
11 of 385
Understanding +4 | -1
A charged particle is released from rest in a region of uniform magnetic field. The path of the particle will be
View answer and explanation
Correct option: D
It will not experience any force
Option D (It will not experience any force) is correct because a charged particle at rest in a magnetic field experiences no magnetic force.
12 Medium
12 of 385
Application +4 | -1
An electron of charge e moves in a circular orbit of radius rr with frequency vv. The magnetic field at the center of orbit is
View answer and explanation
Correct option: C
μ0ev2r\frac{\mu_0 e v}{2 r}
Option C (μ0ev2r\frac{\mu_0 e v}{2 r}) is correct because it is calculated using the formula for the magnetic field at the center of a circular loop.
13 Medium
13 of 385
Application +4 | -1
A long straight wire of radius a carries a steady current I.\text{I}. the current is uniformly distributed over its cross section. The ratio of the magnetic fields B\text{B} and B′,\mathrm{B'}, at radial distances a2\frac{\mathrm{a}}{2} and 2a\text{2a} respectively, from the axis of the wire is:
View answer and explanation
Correct option: C
1
Option C (1) is correct because the ratio of magnetic fields is calculated using Ampere's law, considering the current distribution inside and outside the wire.
14 Medium
14 of 385
Application +4 | -1
A solid cylindrical rod of cross-sectional radius R=2  cmR=2 \,{~cm} is carrying current II along its axis. The magnetic field at a perpendicular distance r=1  cmr=1 \,{~cm} from the central axis is
View answer and explanation
Correct option: C
μ0I8π\frac{\mu_0 I}{8 \pi}
Option D (μ0I8π\frac{\mu_0 I}{8 \pi}) is correct because the magnetic field inside the cylindrical rod is proportional to the distance from the center.
15 Medium
15 of 385
Analyzing +4 | -1
A charged particle is whirled in a horizontal circle on a frictionless table by attaching it to a string fixed at one end. If a magnetic field is switched on in the vertical direction, the tension in the string
View answer and explanation
Correct option: D
May increase or decrease
Option D (May increase or decrease) is correct because the magnetic force can either increase or decrease the tension depending on the direction of the magnetic field and the charge of the particle.

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16 Medium
16 of 385
Application +4 | -1
The magnetic field at a centre of solenoid is BB. The solenoid is cut into two equal parts. For the same current, what is BB at the centre of new solenoid ?
View answer and explanation
Correct option: A
B\mathrm{B}
Option D (4.2×10−5 m/s;2.1×10−5 m/s4.2 \times 10^{-5} \mathrm{~m/s}; 2.1 \times 10^{-5}\mathrm{~m/s}) is correct because it is calculated by applying conservation of momentum and conservation of energy.
17 Medium
17 of 385
Application +4 | -1
The coil of a galvanometer is 0.02 m×0.08 m0.02 \mathrm{~m} \times 0.08 \mathrm{~m} It consists of 200 turns of fine wire and is in a magnetic field of 0.2 tesla. The restoring torque constant of the suspension fibre is 10−6Nm deg−110^{-6} \mathrm{Nm} \mathrm{~deg}^{-1}. Assuming the magnetic field to be radial, (a) what is the maximum current that can be measured by this galvanometer, if the scale can accommodate 30∘30^{\circ} deflection ? (b) What is the smallest current that can be detected, if the minimum observable deflection is 0.1 degree?
View answer and explanation
Correct option: A
4.69×10−4 A;  1.56×10−6 A4.69 \times 10^{-4} \mathrm{~A};\,\,1.56\times10^{-6}\mathrm{~A}
Option A (4.69×10−4 A;,1.56×10−6 A4.69 \times 10^{-4} \mathrm{~A};\\,1.56\times10^{-6}\mathrm{~A}) is correct because it correctly calculates the maximum and minimum current based on the given parameters and the galvanometer's properties.
18 Medium
18 of 385
Understanding +4 | -1
Assertion (A): For a charged particle describing circular path in a uniform transverse magnetic field, the kinetic energy remain constant.

Reason (R): The work done by the force due to magnetic field on a moving charge is always zero.
View answer and explanation
Correct option: A
(A) and (R) are true and (R) is the correct explanation of (A)
Option A ((A) and (R) are true and (R) is the correct explanation of (A)) is correct because the magnetic force does no work on a moving charge, so its kinetic energy remains constant.
19 Medium
19 of 385
Understanding +4 | -1
The resistance of an ideal galvanometer must be:
View answer and explanation
Correct option: D
Zero
Option D (Zero) is correct because an ideal galvanometer should have zero resistance to avoid affecting the circuit's current.
20 Medium
20 of 385
Application +4 | -1
An electron of mass me{m_e} initially at rest moves through a certain distance in a uniform electric field in time t1.{t_1}. A proton of mass mp{m_p} also initially at rest takes time t2{t_2} to move through and equal distance in this uniform electric field. Neglecting the effect of gravity, the ratio of t2/t1{t_2}/{t_1} is nearly equal to
View answer and explanation
Correct option: B
(mp/me)1/2{\left( {{m_p}/{m_e}} \right)^{1/2}}
Option B ((mp/me)1/2{\left( {{m_p}/{m_e}} \right)^{1/2}}) is correct because the ratio of times is derived from the relationship between distance, acceleration, and mass in a uniform electric field.
21 Medium
21 of 385
Analyzing +4 | -1
Assertion: In external magnetic field, kinetic energy of charge particle remains constant.
Reason: In external magnetic field velocity of charge remains constant.
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 because the kinetic energy remains constant in a magnetic field, but the velocity's direction changes, not its magnitude.
22 Medium
22 of 385
Application +4 | -1
An equilateral Δ\Delta, square and a circular loop are made by wire of same length ll if current is same then compare their magnet moment.
View answer and explanation
Correct option: A
l24πi\frac{l^2}{4 \pi} i
Option A (l24πi\frac{l^2}{4 \pi} i) is correct because the magnetic moment of a current loop is given by the product of the current and the area of the loop. The circular loop has the largest area for a given perimeter.
23 Medium
23 of 385
Understanding +4 | -1
The work done by a magnetic field, on a moving charge is
View answer and explanation
Correct option: C
zero because F→\overrightarrow{\mathrm{F}} acts perpendicular to v→\overrightarrow{\mathrm{v}}
Option C (zero because F→\overrightarrow{\mathrm{F}} acts perpendicular to v→\overrightarrow{\mathrm{v}}) is correct because the magnetic force is always perpendicular to the velocity, resulting in zero work done.
24 Medium
24 of 385
Application +4 | -1
A particle with charge 7.00μC7.00 \mu \mathrm{C} is moving with velocity v⃗=−(4×103 m s−1)j^\vec{v}=-\left(4 \times 10^3 \mathrm{~m} \mathrm{~s}^{-1}\right) \hat{j}. The magnetic force on the particle is measured to be F⃗=+(8.4×10−2 N)i^−(5.60×10−2 N)k^\vec{F}=+\left(8.4 \times 10^{-2} \mathrm{~N}\right) \hat{i}-\left(5.60 \times 10^{-2} \mathrm{~N}\right) \hat{k}.
(a) Calculate all the components of the magnetic field you can from this information.
(b) Calculate the scalar product B⃗⋅F⃗\vec{B} \cdot \vec{F}. What is the angle between B⃗\vec{B} and F⃗\vec{F}?
View answer and explanation
Correct option: A
−2 T,−3 T;0,90∘-2 \mathrm{~T}, -3 \mathrm{~T} ; 0, 90^{\circ}
Option A (−2 T,−3 T;0,90∘-2 \text{{ T}}, -3 \text{{ T}}; 0, 90^{\circ}) is correct because the magnetic field components are calculated using the Lorentz force equation, and the scalar product of B and F is zero, indicating they are perpendicular.
25 Medium
25 of 385
Application +4 | -1
An observer's diary reads as follows; "a charged particle is projected in a magnetic field of and particle's acceleration is found to be (xi^+7.0j^)×10−6 m/(x \hat{i}+7.0 \hat{j}) \times 10^{-6} \mathrm{~m} / s2\mathbf{s}^2. Find the value of x\mathbf{x}.
View answer and explanation
Correct option: A
x=3.0x=3.0
Option A (x=3.0x=3.0) is correct because the force due to the magnetic field is perpendicular to the magnetic field itself, allowing us to find x.
26 Medium
26 of 385
Application +4 | -1
A long, cylindrical iron core of cross-sectional area 5.00 cm25.00 \mathrm{~cm}^2 is inserted into a long solenoid having 2000 turns m−1m^{-1} and carrying a current 2.00 A2.00 \mathrm{~A}. The magnetic field inside the core is found to be 1.57 T1.57 \mathrm{~T}. Neglecting the end effects, find the magnetization I of the core and the pole strength developed.
View answer and explanation
Correct option: B
1.25×106 A m−1;  625 A m1.25 \times 10^6 \mathrm{~A~m}^{-1};\,\,625 \mathrm{~A~m}
Option B (1.25×106 A m−1;  625 A m1.25 \times 10^6 \mathrm{~A~m}^{-1};\,\,625 \mathrm{~A~m}) is correct because it correctly calculates the magnetization and pole strength using the given magnetic field and solenoid parameters.
27 Medium
27 of 385
Application +4 | -1
An electron of mass mm is accelerated by a d.c. potential difference VV and then subjected to a transverse magnetic field BB, The trajectory of the electron will be a circle of radius
View answer and explanation
Correct option: D
2mVeB2\sqrt{\frac{2 m V}{eB^2}}
Option D (2mVeB2\sqrt{\frac{{2 m V}}{{eB^2}}}) is correct because it is derived from equating the kinetic energy gained by the electron to its potential energy and then using the formula for the radius of a charged particle in a magnetic field.
28 Medium
28 of 385
Understanding +4 | -1
Assertion: Free electron always keep on moving in a conductor even then no magnetic force act on them in magnetic field unless a current is passed through it.

Reason: The average velocity of free electron is zero.
View answer and explanation
Correct option: A
(A) and (R) are true and (R) is the correct explanation of (A)
Option A ((A) and (R) are true and (R) is the correct explanation of (A)) is correct because free electrons are always in motion, but their average velocity is zero, resulting in no net magnetic force unless a current is present.
29 Medium
29 of 385
Application +4 | -1
A wire of length ' LL ' carries current ' II ' along xx axis, A magnetic field B=B0(i^−j^−k^)TB=B_0(\hat{\mathbf{i}}-\hat{\mathbf{j}}-\hat{\mathbf{k}}) \mathbf{T} acts on the wire. The magnitude of magnetic force acting on the wire is
View answer and explanation
Correct option: C
2ILB0\sqrt{2} \mathrm{ILB}_0
Option C (2ILB0\sqrt{2} \mathrm{ILB}_0) is correct because it's the magnitude of the cross product between the current element and the magnetic field.
30 Medium
30 of 385
Understanding +4 | -1
Assertion: The energy of charged particle moving in a uniform magnetic field does not change.

Reason: Work done by magnetic field on the charge is zero.
View answer and explanation
Correct option: A
(A) and (R) are true and (R) is the correct explanation of (A)
Option A ((A) and (R) are true and (R) is the correct explanation of (A)) is correct because the magnetic force is always perpendicular to the velocity, resulting in zero work done and thus no change in energy.
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