Skip to main content
Back to MAGNETIC EFFECT OF CURRENT

Your Progress

0 / 385 Answered
0
Correct
0
Incorrect
0%
Accuracy
JEE • PHYSICS • MAGNETIC EFFECT OF CURRENT

Magnetic effect of current practice · Medium

Choose an answer to check it, or open “View answer” to study the explanation. Sign in to save progress to your account.

Practice format and chapter quiz

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.

Take a chapter quiz →
Showing 91-120 of 385 questions

Preparing practice and saved progress…

91 Medium
91 of 385
Application +4 | -1
If the maximum value of accelerating potentia provided by a radio frequency oscillator be 20 kV20\,\mathrm{kV}, find the number of revolutions made by a proton in a cyclotron ts achieve one fifth of the speed of light. Mass of a proton =1.67×10−27 kg=1.67 \times 10^{-27} \mathrm{~kg}.
View answer and explanation
Correct option: A
470
Option A (470) is correct because it is the number of revolutions calculated using the kinetic energy gained by the proton in the cyclotron.
92 Medium
92 of 385
Application +4 | -1
A particle of mass mm and charge qq enters a magnetic field BB perpendicularly with a velocity vv. The radius of the circular path described by it will be
View answer and explanation
Correct option: D
mv/Bqmv/Bq
Option D (mv/Bqmv/Bq) is correct because it is derived from equating the magnetic force on the charge to the centripetal force required for circular motion.
93 Medium
93 of 385
Understanding +4 | -1
An arbitrary shaped closed coil is made of a
wire of length L and a current I ampere is
flowing through it. The plane of the coil is
perpendicular to magnetic field B, the force on
the coil is
View answer and explanation
Correct option: A
zero
Option A (zero) is correct because the net force on a closed current loop in a uniform magnetic field is always zero.
94 Medium
94 of 385
Analyzing +4 | -1
Assertion (A) : A charged particle enters in a magnetic field B=B0i^\mathbf{B}=B_0 \hat{\mathbf{i}} with velocity v=v0i^+v0j^\mathbf{v}=v_0 \hat{\mathbf{i}}+v_0 \hat{\mathbf{j}}, then minimum speed of charged particle may be v0v_0.

Reason (R) : A variable acceleration will act on the charged particle.
View answer and explanation
Correct option: D
If Assertion is false but the Reason is true.
Option D (If Assertion is false but the Reason is true) is correct because the speed remains constant, and the acceleration is variable due to the changing direction of velocity.
95 Medium
95 of 385
Analyzing +4 | -1
A charged particle is projected in to a region where there may have an electric filed E⃗\vec{E} and/or magnetic field Bˉ\bar{B}. If the charged particle goes un-accelerated, then it is not possible that
View answer and explanation
Correct option: B
E⃗≠0,B⃗=0\vec{E} \neq 0, \vec{B}=0
Option B (E⃗≠0,B⃗=0\vec{E} \neq 0, \vec{B}=0) is correct because if only an electric field is present, the charged particle will always accelerate.
96 Medium
96 of 385
Understanding +4 | -1
‘Biot-Savarts’ law of magnetism is analogous
to:
View answer and explanation
Correct option: A
Coulomb’s law
Option A (Coulomb’s law) is correct because both Biot-Savart's law and Coulomb's law describe fields that vary inversely with the square of the distance from the source.
97 Medium
97 of 385
Understanding +4 | -1
A proton enters a magnetic field with velocity
parallel to the magnetic field. The path
followed by the proton is a
View answer and explanation
Correct option: D
straight line
Option D (straight line) is correct because the magnetic force on a moving charge is zero when the velocity is parallel to the magnetic field, so the proton continues in a straight line.
98 Medium
98 of 385
Understanding +4 | -1
A charge qq moves with velocity v⃗\vec{v} in the direction of the applied field B⃗\vec{B}. The force that acts on the charge is
View answer and explanation
Correct option: D
F→=0\overrightarrow{\mathrm{F}}=0
The correct answer is F→=0\overrightarrow{\mathrm{F}}=0 because the force on a charge moving parallel to a magnetic field is zero.
99 Medium
99 of 385
Application +4 | -1
An electron and a proton, both having the same kinetic energy, enter a region of uniform magnetic field, in a plane perpendicular to the field. If their masses are denoted by mem_e and mpm_p respectively, then the ratio of the radii (electron to proton) of their circular orbits is
View answer and explanation
Correct option: B
memp\sqrt{\frac{m_e}{m_p}}
Option B (memp\sqrt{\frac{m_e}{m_p}}) is correct because the ratio of radii is proportional to the square root of the mass ratio when kinetic energy and magnetic field are the same.
100 Medium
100 of 385
Application +4 | -1
A galvanometer coil is replaced by another coil of diameter one-fourth of the original diameter and the total number of turns as ten times the original number. What will be the new deflection if the same current is passed through it? Old deflection is θ\theta.
View answer and explanation
Correct option: A
58θ\frac{5}{8} \theta
Option A (58θ\frac{5}{8} \theta) is correct because it is the new deflection calculated based on the changes in diameter and number of turns.
101 Medium
101 of 385
Understanding +4 | -1
Assertion (A): Magnetic field interacts with a moving charge and not with a stationary charge.

Reason (R): Magnetic field will be produced by a moving charge but not by a stationary charge.
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 a moving charge creates a magnetic field, which then interacts with other moving charges.
102 Medium
102 of 385
Remembering +4 | -1
A charge qq is moving with a velocity vv parallel to a magnetic field BB. Force on the charge due to magnetic field is
View answer and explanation
Correct option: C
zero
Option C (zero) is correct because the force on a charge moving parallel to a magnetic field is zero, as the angle between velocity and magnetic field is 0 degrees, and sin⁡(0)=0\sin(0) = 0.
103 Medium
103 of 385
Understanding +4 | -1
A current carrying coil is subjected to a
uniform magnetic field. The coil will orient so
that its plane becomes
View answer and explanation
Correct option: D
perpendicular to the magnetic field
Option D (perpendicular to the magnetic field) is correct because a current loop aligns its magnetic moment with the external field, minimizing energy.
104 Medium
104 of 385
Application +4 | -1
A short bar magnet is placed with its axis at 30∘30^{\circ} with an external field 0.05 T0.05 \mathrm{~T}. If the magnetic moment of the magnet is 0.8Am20.8 \mathrm{Am}^2, then the torque experienced by the magnet is
View answer and explanation
Correct option: D
0.02 N.m0.02 \mathrm{~N} . \mathrm{m}
Option C (250 N250 {~N}) is correct because it is calculated using the impulse-momentum theorem and assuming the force decreases linearly.
105 Medium
105 of 385
Application +4 | -1
A solenoid of length 0.5 m0.5 \mathrm{~m} has a radius of 1 cm1 \mathrm{~cm} and is made up of 500 turns. It carries a current of 5 A5 \mathrm{~A}. What is the magnitude of the magnetic field inside the solenoid?
View answer and explanation
Correct option: A
6.28×10−3T6.28 \times 10^{-3} \mathrm{T}
Option A (6.28×10−3T6.28 \times 10^{-3} \mathrm{T}) is correct because the magnetic field inside a long solenoid is calculated using B=μ0nIB = \mu_0 n I.
106 Medium
106 of 385
Remembering +4 | -1
The gyromagnetic ratio of an electron == ____\_\_\_\_ specific charge of an electron.
View answer and explanation
Correct option: C
12\frac{1}{2}
Option C (12\frac{1}{2}) is correct because the gyromagnetic ratio is half of the specific charge of an electron.
107 Medium
107 of 385
Application +4 | -1
A long solenoid is formed by winding 20 turns/cm. The current necessary to produce a magnetic field of 20 millitesla inside the solenoid will be approximately

(μ04π=10−7 tesla − metre / ampere )\left(\frac{\mu_0}{4 \pi}=10^{-7} \text { tesla }- \text { metre } / \text { ampere }\right)
View answer and explanation
Correct option: A
8.0 A8.0 \mathrm{~A}
Option A (8.0 A) is correct because it accurately calculates the current required to produce the given magnetic field inside the solenoid using the formula B=μ0niB = \mu_0 n i.
108 Medium
108 of 385
Understanding +4 | -1
In a cyclotron, a charged particle
View answer and explanation
Correct option: A
undergoes acceleration all the time
Option A is correct because the charged particle is always experiencing a force, either from the electric field between the dees or the magnetic field within the dees, resulting in continuous acceleration.
109 Medium
109 of 385
Understanding +4 | -1
A current of ii ampere flows along an infinitely long straight thin walled tube, then the magnetic induction at any point inside the tube is
View answer and explanation
Correct option: B
zero
Option B (zero) is correct because according to Ampere's Law, the magnetic field inside a hollow current-carrying conductor is zero.
110 Medium
110 of 385
Understanding +4 | -1
Assertion: A planar circular loop of area AA and current II is equivalent to magnetic dipole of dipole moment M=IA.M = IA.
Reason: At large distances, magnetic field of circular loop and magnetic dipole is same.
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 a current loop behaves as a magnetic dipole, and their fields are similar at large distances.
111 Medium
111 of 385
Understanding +4 | -1
A current carrying small loop behaves like a small magnet. If AA be its area and MM its magnetic moment, the current in the loop will be
View answer and explanation
Correct option: A
M/AM/A
Option A (M/AM/A) is correct because the magnetic moment of a current loop is given by M=iAM = iA, so i=M/Ai = M/A.
112 Medium
112 of 385
Understanding +4 | -1
If an ammeter is to be used in place of a voltmeter, then we must connect with the ammeter a
View answer and explanation
Correct option: C
High resistance in series
Option C (High resistance in series) is correct because to use an ammeter as a voltmeter, you need to increase its resistance significantly, which is achieved by adding a high resistance in series.
113 Medium
113 of 385
Application +4 | -1
An electron having mass 9×10−31 kg9 \times {10^{ - 31}}\,kg, charge 1.6×10−19C1.6 \times {10^{ - 19}}C and moving with a velocity of 106 m/s{10^6}\,m/s enters a region where magnetic field exists. It is describes a circle of radius 0.10 m, the intensity of magnetic field must be [in Tesla]
View answer and explanation
Correct option: B
5.6×10−55.6 \times {10^{ - 5}}
Option B (5.6×10−55.6 \times {10^{ - 5}}) is correct because it is the magnetic field intensity calculated using the formula for the radius of a circular path of a charged particle in a magnetic field.
114 Medium
114 of 385
Remembering +4 | -1
Magnetic field at any point on the axis of a current element is
View answer and explanation
Correct option: A
Zero
Option A (Zero) is correct because the magnetic field at a point on the axis of a current element is zero due to the angle in the Biot-Savart Law.
115 Medium
115 of 385
Understanding +4 | -1
Similar or same magnetic fields can be
produced by
View answer and explanation
Correct option: A
a solenoid and a bar magnet
Option A (a solenoid and a bar magnet) is correct because both produce similar magnetic field patterns, with strong and uniform fields inside and stronger fields at the poles.
116 Medium
116 of 385
Application +4 | -1
If we double the radius of a coil keeping the current through it unchanged, then the magnetic field at any point at a large distance from the centre becomes approximately
View answer and explanation
Correct option: C
four times
Option C (four times) is correct because the magnetic field at a large distance is proportional to the square of the radius.
117 Medium
117 of 385
Understanding +4 | -1
The current sensitivity of a moving coil galvanometer depends on
View answer and explanation
Correct option: A
the number of turns in the coil
Option A is correct because the current sensitivity of a moving coil galvanometer is directly proportional to the number of turns in the coil.
118 Medium
118 of 385
Understanding +4 | -1
A particle of mass mm, charge QQ and kinetic energy is 10 J10 \mathrm{~J} enters a transverse uniform magnetic field B⃗\vec{B}. After 5 seconds the kinetic energy of the particle will be :-
View answer and explanation
Correct option: B
10 J
Option B is correct because a magnetic field only changes the direction of a moving charge, not its speed, so the kinetic energy remains constant.
119 Medium
119 of 385
Application +4 | -1
Proton (charge +e+\mathrm{e}, mass m\mathrm{m} ) are being accelerated in a cyclotron where a magnetic field B⃗\vec{B} is applied perpendicular to the dees and a battery of VV volts in connected across the dees to accelerate the protons. The number of revolutions per second (frequency in Hz\mathrm{Hz} ) by the protons is
View answer and explanation
Correct option: B
eB2πm\frac{\mathrm{eB}}{2 \pi \mathrm{m}}
Option B (eB2πm\frac{{\mathrm{{eB}}}}{{{{2 \pi \mathrm{{m}}}}}}) is correct because it represents the cyclotron frequency, which depends on the charge, magnetic field, and mass of the proton.
120 Medium
120 of 385
Application +4 | -1
A current carrying loop of radius rr carries a current II in clock wise direction. It is placed in a uniform magnetic field BB such that plane of the coil is parallel to field. The net torque on the coil is
View answer and explanation
Correct option: A
πr2IB\pi r^2 I B
Option A (πr2IB\pi r^2 I B) is correct because the torque on a current loop in a magnetic field is given by τ⃗=M⃗×B⃗\vec{\tau} = \vec{M} \times \vec{B}, where M is the magnetic moment.
Revision notes and key takeaways

Review MAGNETIC EFFECT OF CURRENT in your textbook or class notes, then use this set to check your understanding.

Showing 91-120 of 385 questions

Your practice

0 / 385 answered · 385 remaining · 0% accuracy

Progress is stored on this device. Sign in to save to your account.

✓ Correct · ✗ Incorrect · ○ Unanswered

Answer feedback and difficulty

Type at least 2 characters or jump directly:
🔓

Save your learning progress

Sign in with a free account to save your practice answers and quiz results, then return to them when you revise.