Skip to main content
Back to MAGNETIC EFFECT OF CURRENT

Your Progress

0 / 418 Answered
0
Correct
0
Incorrect
0%
Accuracy
JEE • PHYSICS • MAGNETIC EFFECT OF CURRENT

Magnetic effect of current practice · Mixed difficulty

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 211-240 of 418 questions

Preparing practice and saved progress…

211 Medium
211 of 418
Analyzing +4 | -1
Assertion: Magnetic force between two charges is generally much smaller than the electric force between them.
Reason: Speeds of charges are much smaller than the free–space speed of light.
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 the magnetic force is much smaller than the electric force due to the relatively low speeds of charges compared to the speed of light.
212 Easy
212 of 418
Understanding +4 | -1
In which of the following case no force exerted
by a magnetic field on a charge ?
View answer and explanation
Correct option: C
At rest
Option C is correct because a magnetic field exerts no force on a charge at rest since the magnetic force is proportional to the velocity of the charge.
213 Medium
213 of 418
Application +4 | -1
A proton enters into a magnetic field of induction 1.732 T1.732 \mathrm{~T}, with a velocity of 107 m/s10^7 \mathrm{~m} / \mathrm{s} at an angle 60∘60^{\circ} to the field. The force acting on the proton is (e=1.6×10−19C,sin⁡60∘=cos⁡30∘=\left(\mathrm{e}=1.6 \times 10^{-19} \mathrm{C}, \sin 60^{\circ}=\cos 30^{\circ}=\right. 32\frac{\sqrt{3}}{2} )
View answer and explanation
Correct option: D
2.4×10−12 N2.4 \times 10^{-12} \mathrm{~N}
Option D (2.4×10−12 N2.4 \times 10^{-12} \mathrm{~N}) is correct because the force acting on the proton is calculated using the formula F=qvBsin⁡θF = qvB \sin{\theta}.
214 Medium
214 of 418
Application +4 | -1
A long straight wire carries a current 15 A15 \mathrm{~A}. Calculate the magnetising field H\mathrm{H} at a point at distance 0.105 m0.105 \mathrm{~m} from the axis of wire
View answer and explanation
Correct option: C
22.7 A/m22.7 \mathrm{~A} / \mathrm{m}
Option C (22.7 A/m) is correct because it is calculated using the formula for the magnetic field around a long straight wire.
215 Hard
215 of 418
Application +4 | -1
A solenoid of 1000 turns per meter has a core with relative permeability 500. Insulated windings of the solenoid carry an electric current of 5A. The magnetic flux density produced by the solenoid is (permeability of free space =  4π×10−7H/m= \,\,4\pi \times {10^{ - 7}}\mathrm{H/m} )
View answer and explanation
Correct option: A
πT\pi T
Option A (πT\pi T) is correct. The magnetic flux density is calculated using the formula B = μ_r μ_0 n I, where μ_r is the relative permeability, μ_0 is the permeability of free space, n is the number of turns per meter, and I is the current.
216 Medium
216 of 418
Application +4 | -1
A circular coil carrying a certain current produces a magnetic field B0B_0 at its centre. The coil is now rewound so as to have 4 turns and the same current is passed through it. The new magnetic field at the centre is
View answer and explanation
Correct option: B
16B016 B_0
Option A (245.01 V/m245.01 \mathrm{~V} / \mathrm{m}) is correct because it is the amplitude of the electric field of radiation at the given distance from the lamp, calculated using the Poynting vector and the relationship between electric and magnetic fields.
217 Medium
217 of 418
Analyzing +4 | -1
Statement-I: If by mistake a voltmeter is connected in series it gets burnt
Statement-II: Current will drastically decrease in the circuit.
View answer and explanation
Correct option: D
Statement-I is false, but Statement-II is true
Option D (Statement-I is false, but Statement-II is true) is correct because a voltmeter has high resistance, so connecting it in series drastically decreases the current, but it doesn't necessarily burn out.
218 Medium
218 of 418
Analyzing +4 | -1
The turns of a solenoid, designed to provide a given magnetic
flux density along its axis, are wound to fill the space between
two concentric cylinders of fixed radii. How should the diameter
dd of the wire used be chosen so as to minimize the heat
dissipated in the windings?
View answer and explanation
Correct option: C
Wire should be multiple of dd
Option C (Wire should be multiple of dd) is correct because the heat dissipated is independent of the diameter of the wire, so any multiple of d will work, but the most practical is a multiple of d itself.
219 Medium
219 of 418
Understanding +4 | -1
Fleming's left and right hand rules are used in:
View answer and explanation
Correct option: C
DC motor and DC generator
Option C (DC motor and DC generator) is correct because Fleming's left-hand rule is used for DC motors, and Fleming's right-hand rule is used for DC generators.
220 Easy
220 of 418
Application +4 | -1
The ratio of time periods of α\alpha-particle and proton moving on circular path in uniform transverse magnetic field :-
View answer and explanation
Correct option: A
2:12: 1
Option A (2:1) is correct because the time period is proportional to the mass and inversely proportional to the charge, and an alpha particle has twice the mass and twice the charge of a proton.
221 Medium
221 of 418
Remembering +4 | -1
The magnetic induction at the centre of a current carrying circle of radius rr, is
View answer and explanation
Correct option: B
Inversely proportional to rr
Option B (Inversely proportional to rr) is correct because the magnetic field at the center of a current-carrying circle is inversely proportional to the radius.
222 Medium
222 of 418
Application +4 | -1
A charge of 3.14×10−6C3.14 \times 10^{-6} \mathrm{C} is distributed uniformly over a circular ring of radius 20.0 cm20.0 \mathrm{~cm}. The ring rotates about its axis with an angular velocity of 60.0 rad s−160.0\, \mathrm{rad}\, \mathrm{s}^{-1}. Find the ratio of the electric field to the magnetic field at a point on the axis at a distance of 5.00 cm5.00 \mathrm{~cm} from the centre.
View answer and explanation
Correct option: D
38\frac{3}{8}
Option D (38\frac{3}{8}) is correct because it is the ratio of the electric field to the magnetic field calculated using the given parameters and formulas.
223 Medium
223 of 418
Understanding +4 | -1
Two parallel wires are carrying electric currents of equal magnitude and in the same direction. They exert:
View answer and explanation
Correct option: A
an attractive force on each other
Option A (an attractive force on each other) is correct because parallel currents attract each other due to the magnetic fields they create.
224 Medium
224 of 418
Application +4 | -1
A solenoid 60 cm60 \mathrm{~cm} long and of radius 4.0 cm4.0 \mathrm{~cm} has 3 layers of windings of 300 turns each. A 2.0 cm2.0 \mathrm{~cm} long wire of mass 2.5 g2.5 \mathrm{~g} lies inside the solenoid near its centre normal to its axis; both the wire and the axis of the solenoid are in the horizontal plane. The wire is connected through two leads parallel to the axis of the solenoid to an external battery which supplies a current of 6.0 A6.0 \mathrm{~A} in the wire. What value of current in the windings of the solenoid can support the weight of the wire ? g=9.8 ms−2g=9.8 \mathrm{~ms}^{-2}.
View answer and explanation
Correct option: A
108.3 A
Option A (108.3 A) is correct because the current required to support the wire is calculated by equating the magnetic force on the wire to its weight.
225 Hard
225 of 418
Application +4 | -1
An ideal coil of 10 henry is joined in series with a resistance of 5ohm5 \mathrm{ohm} and a battery of 5 volt. After two seconds joining, the current flowing in ampere in the circuit will be :-
View answer and explanation
Correct option: B
(1−e−1)\left(1-\mathrm{e}^{-1}\right)
Option B (1−e−11-\mathrm{e}^{-1}) is correct because it represents the current flowing in the L-R circuit after two seconds, calculated using the formula for current growth in an LR circuit.
226 Medium
226 of 418
Application +4 | -1
An electron having charge ' ee ' is moving with a constant speed vv along a circle of radius rr. Its magnetic moment will be
View answer and explanation
Correct option: B
evr2\frac{\mathrm{e} v \mathrm{r}}{2}
Option B (evr2\frac{\mathrm{e} v \mathrm{r}}{2}) is correct because the magnetic moment is proportional to the current and the area of the loop.
227 Medium
227 of 418
Application +4 | -1
A 0.8 m long solenoid has 800 turns and has a field density of 2.52×10−3T2.52\mathrm{ } \times \mathrm{ }{10^{ - 3}}T at its centre. What is current in the wire ?
View answer and explanation
Correct option: B
2
Option B (2) is correct because the current is calculated using the formula for the magnetic field inside a solenoid.
228 Medium
228 of 418
Application +4 | -1
A galvanometer of resistance 25  Ω25\,\,\mathrm{\Omega } giving full scale deflection for a current of 10  milli  ampere10\,\,milli\,\,ampere, is to be changed into a voltmeter of range 100  V100\,\,V by connecting a resistance of ′R′'R' in series with galvanometer. The value of resistance RR in Ω\mathrm{\Omega } is
View answer and explanation
Correct option: D
9975
Option D (9975) is correct because the series resistance needed to convert a galvanometer into a voltmeter is calculated using the formula R = V/ig - G.
229 Medium
229 of 418
Application +4 | -1
A galvanometer has resistance 500 ohm. It is shunted so that its sensitivity decreases by 100 times. Find the shunt resistance.
View answer and explanation
Correct option: C
5.05 Ω5.05\,\Omega
Option C (5.05 Ω5.05\,\Omega) is correct because the shunt resistance is calculated using the formula S=Gn−1S = \frac{G}{{n - 1}}, where G is the galvanometer resistance and n is the factor by which sensitivity decreases.
230 Medium
230 of 418
Application +4 | -1
Two long straight conductors carry currents 3A3 A and 1A1 A into the plane of paper. A circular path is imagined to enclose these currents. The value of ∮B⃗⋅dl→\oint \vec{B} \cdot \overrightarrow{d l} is
View answer and explanation
Correct option: B
4μ04 \mu_0
Option B (4μ04 \mu_0) is correct because Ampere's law states that the line integral of the magnetic field around a closed loop is proportional to the enclosed current.
231 Medium
231 of 418
Application +4 | -1
A galvanometer with a coil resistance of 100  Ω100\,\,\Omega gives a full-scale deflection when a current of 1  mA1\,\,mA is passed through it. The resistance of the shunt needed in ohm to convert this galvanometer into an ammeter of range 10 A10\,A is nearly.
View answer and explanation
Correct option: A
0.01
Option D (0.099) is correct because it is calculated using the formula for shunt resistance in an ammeter.
232 Medium
232 of 418
Understanding +4 | -1
When a charged particle enters a uniform magnetic field
View answer and explanation
Correct option: C
Speed of particle will be constant
Option C (Speed of particle will be constant) is correct because the magnetic force is always perpendicular to the velocity, doing no work and thus not changing the speed.
233 Medium
233 of 418
Understanding +4 | -1
Armature current is maximum in a DC motor when the motor has
View answer and explanation
Correct option: B
just started
Option B (just started) is correct because at the moment the DC motor starts, the back EMF is zero, resulting in maximum armature current.
234 Medium
234 of 418
Application +4 | -1
A galvanometer of resistance 20  Ω\mathrm{20}\,\,\mathrm{\Omega } shows a deflection of 10\mathrm{10} divisions when a current of 1  mA\mathrm{1}\,\,\mathrm{mA} is passed through it. If a shunt of 4  Ω\mathrm{4}\,\,\mathrm{\Omega } is connected and there are 50\mathrm{50} divisions on the scale, the range of the galvanometer is
View answer and explanation
Correct option: D
30  mA\mathrm{30}\,\,\mathrm{mA}
Option D (30 mA\mathrm{30}\,\mathrm{mA}) is correct because it is the calculated range of the galvanometer after connecting the shunt resistance.
235 Medium
235 of 418
Understanding +4 | -1
A charged particle enters in a uniform
magnetic field with a certain velocity. The
power delivered to the particle by the magnetic
field depends on
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the magnetic force does no work on a charged particle, and therefore the power delivered by the magnetic field is zero.
236 Medium
236 of 418
Understanding +4 | -1
Assertion : In electric circuits, wires carrying
currents in opposite directions are often twisted
together.
Reason: If the wires are not twisted together, the
combination of the wires forms a current loop, the
magnetic field generated by the loop might affect
adjacent circuits or components
View answer and explanation
Correct option: A
If both Assertion and Reason are correct and the
Reason is a correct explanation of the Assertion
Option A is correct because twisting wires carrying current in opposite directions minimizes the magnetic field they produce, preventing interference with nearby circuits.
237 Medium
237 of 418
Application +4 | -1
An alternating electric field, of frequency vv, is applied across the dees (radius =R=R) of a cyclotron that is being used to accelerate protons (mass = mm). The operating magnetic field (B)(B) used in the cyclotron and the kinetic energy (KK) of the proton beam, produced by it, are given by
View answer and explanation
Correct option: A
B=2πmveB=\frac{2 \pi m v}{e} and K=2mπ2v2R2K=2 m \pi^2 v^2 R^2
Option A (B=2πmveB=\frac{2 \pi m v}{e} and K=2mπ2v2R2K=2 m \pi^2 v^2 R^2) is correct because the magnetic field and kinetic energy are derived from the cyclotron frequency and radius equations.
238 Medium
238 of 418
Understanding +4 | -1
Maximum kinetic energy gained by the
charged particle in the cyclotron in
independent of
View answer and explanation
Correct option: D
frequency of revolution
Option D (frequency of revolution) is correct because the maximum kinetic energy in a cyclotron depends on the charge, magnetic field, and radius of the dees, but not directly on the frequency.
239 Medium
239 of 418
Understanding +4 | -1
The work done by an uniform 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 on a moving charge is always perpendicular to its velocity, meaning no work is done.
240 Medium
240 of 418
Analyzing +4 | -1
A straight long conductor carries current along the positive xx-axis. Identify the correct statement related to the four points A(a,a,0),B(a,0,a),C(a,−a,0)A(a, a, 0), B(a, 0, a), C(a,-a, 0) and D(a,0,−a)D(a, 0,-a).
View answer and explanation
Correct option: D
All of the above
Option D is correct because the magnetic field's magnitude is the same, fields at A and B are perpendicular, and fields at A and C are antiparallel due to the geometry around the current-carrying wire.
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 211-240 of 418 questions

Your practice

0 / 418 answered · 418 remaining · 0% accuracy

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

✓ Correct · ✗ Incorrect · ○ Unanswered

Practice settings

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.