Skip to main content
Back to MAGNETIC EFFECT OF CURRENT

Your Progress

0 / 280 Answered
0
Correct
0
Incorrect
0%
Accuracy
NEET • 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

The practice player uses single-correct-answer MCQs. Chapter quizzes help you review a topic; they are not full-length NEET mock papers.

Take a chapter quiz →
Showing 31-60 of 280 questions

Preparing practice and saved progress…

31 Medium
31 of 280
Understanding +4 | -1
The direction of magnetic field dB⃗d \vec{B} due to a current element Idl⃗\mathrm{I} d \vec{l} at a point of distance rr from it, when a current I passes through a long conductor is in the direction
View answer and explanation
Correct option: C
perpendicular to both dl⃗\mathrm{d} \vec{l} and r→\overrightarrow{\mathrm{r}}
Option C is correct because the magnetic field due to a current element is perpendicular to both the current element and the position vector, as described by the Biot-Savart law.
32 Hard
32 of 280
Application +4 | -1
A charged particle moving along +x+{x} direction with velocity v{v} enters a region where there is a uniform magnetic field B⃗=−Bk^\vec{B}=-B \hat{k} from x=0x=0 to x=dx=d, the particle gets deflected at an angle θ\theta from its intial path. The specific charge of the particle is :
View answer and explanation
Correct option: D
Vsin⁡θBd\frac{{V} \sin \theta}{{Bd}}
Option D (13πqfl2\frac{1}{3} \pi q f l^2) is correct because the magnetic moment is calculated by integrating the current due to each infinitesimal charge element along the rod.
33 Medium
33 of 280
Application +4 | -1
An electron of mass m and charge q is travelling with a speed v along a circular path of radius r at right angles to a uniform magnetic filed B. If speed of the electron is double and the magnetic field is halved, then resulting path would have a radius of
View answer and explanation
Correct option: D
4 r
Option D (4 r) is correct because the radius of the circular path is directly proportional to the velocity and inversely proportional to the magnetic field.
34 Medium
34 of 280
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 is correct. The energy of a charged particle moving in a uniform magnetic field remains constant because the magnetic force does no work on the charge.
35 Easy
35 of 280
Application +4 | -1
A solenoid of length 0.4 m and having 500 turns of wire carries a current of 3 A. A thin coil having 10 turns of wire and of radius 0.01 m carries of 0.4 A. Assuming long solenoid, magnetic field at the centre of the solenoid is
View answer and explanation
Correct option: A
1.5π×10−7 T1.5\pi \times {10^{ - 7}}\,T
Option A (1.5π×10−7 T1.5\pi \times {10^{ - 7}}\,T) is incorrect. The correct answer should be 1.57×10−3T1.57 \times 10^{-3} T. The provided answer seems to have an error in the exponent. The question is asking for the magnetic field at the center of the solenoid, not the field due to the thin coil.
36 Medium
36 of 280
Understanding +4 | -1
The condition under which a point charge moving through a magnetic field, experiences maximum force (in terms of magnitude) is
View answer and explanation
Correct option: C
v→⊥B→\overrightarrow{\mathrm{v}} \perp \overrightarrow{\mathrm{B}}
Option C (v→⊥B→\overrightarrow{\mathrm{v}} \perp \overrightarrow{\mathrm{B}}) is correct because the magnetic force is maximum when the velocity and magnetic field are perpendicular, making the angle 90 degrees and sin(90) = 1.
37 Medium
37 of 280
Remembering +4 | -1
Before using the tangent galvanometer, its coil
is set up in
View answer and explanation
Correct option: A
magnetic meridian
Option A (magnetic meridian) is correct because the coil of a tangent galvanometer must be aligned with the Earth's magnetic meridian for accurate measurements.
38 Medium
38 of 280
Analyzing +4 | -1
Two identical current carrying coaxial loops, carry current ii in an opposite sense. A simple amperian loop passes through both of them once. Calling the loop as CC,
View answer and explanation
Correct option: C
there may be a point on CC where B{B} and dld {l} are perpendicular.
Option C (there may be a point on CC where B{B} and dld {l} are perpendicular) is correct because the magnetic fields from the two loops can cancel each other out at certain points, resulting in perpendicularity.
39 Medium
39 of 280
Application +4 | -1
A proton moving with a velocity of 2×106 ms−1\mathrm{2} \times {10^6}\,\mathrm{m}{\mathrm{s}^{ - 1}} describes a circle of radius RR in a magnetic field. The speed of an α−\alpha - particle to describe a circle of same radius in the same magnetic field is
View answer and explanation
Correct option: A
1×106 m/s1 \times {10^6}\,m/s
Option A (1×106 m/s1 \times {10^6}\,m/s) is correct because the radius of the circular path depends on the charge-to-mass ratio and velocity, and the alpha particle has twice the charge and four times the mass of a proton.
40 Medium
40 of 280
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 r.
41 Hard
41 of 280
Application +4 | -1
A charged particle is fired at an angle θ\theta to a uniform magnetic field directed along the X\mathrm{X}-axis. During its motion, the pitch of the helical path is equal to the maximum distance of the particle from the X\mathrm{X}-axis, then
View answer and explanation
Correct option: D
tan⁡θ=π\tan \theta=\pi
Option A (have the end A point towards north) is correct because the solenoid behaves like a bar magnet, and the end where current enters anticlockwise acts as the north pole.
42 Medium
42 of 280
Understanding +4 | -1
There are three voltmeters of the same range but of resistances 10000  Ω,  8000  Ω10000\,\,\mathrm{\Omega },\,\,8000\,\,\mathrm{\Omega } and 4000  Ω4000\,\,\mathrm{\Omega } respectively. The best voltmeter among these is the one whose resistance is
View answer and explanation
Correct option: A
10000  Ω10000\,\,\mathrm{\Omega }
Option A (10000  Ω10000 \,\,\mathrm{\Omega }) is correct because a voltmeter with higher resistance draws less current from the circuit, providing a more accurate reading.
43 Medium
43 of 280
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 the self-inductance of a solenoid is proportional to the square of the number of turns per unit length.
44 Hard
44 of 280
Application +4 | -1
A constant voltage is applied to a series R−L\mathrm{R}-\mathrm{L} circuit by closing the switch. The voltage across inductor (L=2H)(\mathrm{L}=2 \mathrm{H}) is 20 V20 \mathrm{~V} at t=0\mathrm{t}=0 and drops to 5 V5 \mathrm{~V} at 20 ms20 \mathrm{~ms}. The value of R\mathrm{R} in Ω\Omega is:
View answer and explanation
Correct option: C
100ln⁡4Ω100 \ln 4 \Omega
Option C (100ln⁡4Ω100 \ln 4 \Omega) is correct because it's the value of R derived from the exponential decay of voltage across the inductor in an RL circuit.
45 Medium
45 of 280
Understanding +4 | -1
A magnet is suspended lengthwise from a spring and while it oscillates, the magnet moves in and out of the coil C connected to a galvanometer G. Then as the magnet oscillates.
View answer and explanation
Correct option: D
Deflection of G to the left and right has decreasing amplitude
Option D (Deflection of G to the left and right has decreasing amplitude) is correct because the oscillating magnet induces a current in the coil, which opposes the motion of the magnet, leading to damping.
46 Medium
46 of 280
Application +4 | -1
A cylindrical conductor of radius ‘R’ carries a current ‘i’. The value of magnetic field at a point which is R/4 distance inside from the surface is 10 T. Find the value of magnetic field at point which is 4R distance outside from the surface
View answer and explanation
Correct option: B
83T\frac{8}{3}T
Option B (83T\frac{8}{3}T) is correct because it is derived from applying Ampere's law inside and outside the conductor.
47 Medium
47 of 280
Application +4 | -1
A charge q is spread uniformly over an insulated loop of radius r. If it is rotated with an angular velocity ω\omega with respect to normal axis then the magnetic moment of the loop is
View answer and explanation
Correct option: A
12qωr2\frac{1}{2}q\omega {r^2}
Option A (12qωr2\frac{1}{2}q\omega {r^2}) is correct because the magnetic moment is calculated by integrating the current over the area of the loop.
48 Hard
48 of 280
Application +4 | -1
A winding wire which is used to frame a solenoid can bear a maximum 10 A10 \mathrm{~A} current. If length of solenoid is 80 cm80 \mathrm{~cm} and it's cross sectional radius is 3 cm3 \mathrm{~cm} then required length of winding wire is (B=0.2 T)(\mathrm{B}=0.2 \mathrm{~T}) :-
View answer and explanation
Correct option: C
2.4×103 m2.4 \times 10^3 \mathrm{~m}
Option C (2.4×103 m2.4 \times 10^3 \mathrm{~m}) is correct because it is the calculated length of the winding wire based on the given magnetic field, current, and solenoid dimensions.
49 Medium
49 of 280
Application +4 | -1
A milli volt meter of 25 milli volt range is to be converted into an ammeter of 25 ampere range. The value of necessary shunt will be.
View answer and explanation
Correct option: A
0.001
Option A (0.001) is correct because it represents the shunt resistance needed to divert most of the current through the ammeter.
50 Medium
50 of 280
Analyzing +4 | -1
To decrease the range of an ammeter, its resistance need to be increased. An ammeter has resistance R0{R_0} and range II. Which of the following resistance can be connected in series with it to decreases its range to I/nI/n?
View answer and explanation
Correct option: D
None of these
The correct answer is None of these because to decrease the range of an ammeter, a shunt resistance must be connected in parallel, not a resistance in series.
51 Medium
51 of 280
Application +4 | -1
An electric bulb is rated as 200 W. What will be the peak magnetic field at 4 m distance produced by the radiations coming from this bulb? Consider this bulb as a point source with 3.5% efficiency.
View answer and explanation
Correct option: B
1.71×10−8T1.71 \times {10^{ - 8}}T
Option B (1.71×10−8T1.71 \times {10^{ - 8}}T) is correct because the peak magnetic field is calculated from the intensity of the radiation, considering the bulb's efficiency.
52 Medium
52 of 280
Application +4 | -1
A proton and an α−\alpha - particle enter a uniform magnetic field perpendicularly with the same speed. If proton takes 25 μ25\,\mu second to make 5 revolutions, then the time period for the α−\alpha - particle would be
View answer and explanation
Correct option: C
10 μ sec⁡10\,\mu \,\sec
Option C (10 μ sec⁡10\,\mu \,\sec) is correct because the time period is proportional to the mass and inversely proportional to the charge. The alpha particle has twice the charge and four times the mass of the proton.
53 Medium
53 of 280
Understanding +4 | -1
Select the incorrect statement about Lorentz Force.
View answer and explanation
Correct option: D
For a static charge the magnetic force is maximum.
Option D (For a static charge the magnetic force is maximum) is correct because the magnetic force on a static charge is always zero.
54 Medium
54 of 280
Understanding +4 | -1
A charge QQ is moving dldl distance in the magnetic field BB.find the value of work done by BB.
View answer and explanation
Correct option: C
Zero
Option C is correct. The work done by a magnetic field on a moving charge is always zero because the magnetic force is perpendicular to the displacement.
55 Easy
55 of 280
Application +4 | -1
A long solenoid of radius 1 mm has 100 turns per mm. If 1 A current flows in the solenoid, the magnetic field strength at the centre of the solenoid is:
View answer and explanation
Correct option: C
12.56×10−2 T\mathrm{12}\mathrm{.56} \times \mathrm{1}{\mathrm{0}^{ - 2}}\,\mathrm{T}
Option C (12.56×10−2 T\mathrm{12}\mathrm{.56} \times \mathrm{1}{\mathrm{0}^{ - 2}}\,\mathrm{T}) is correct because it is calculated using the formula for the magnetic field at the center of a long solenoid.
56 Hard
56 of 280
Application +4 | -1
A proton is accelerated in a cyclotron where the applied magnetic field is 0.5 T0.5 {~T}. If the potential gap is 10 kV10 \,{kV}, then how much revolutions the proton has to make between the dee's to acquire a kinetic energy of 12 MeV12 \,{MeV} ?
View answer and explanation
Correct option: A
600
Option A (600) is correct because the number of revolutions is calculated by dividing the total kinetic energy by the energy gained per revolution.
57 Medium
57 of 280
Remembering +4 | -1
The direction of the force on a current carrying
conductor held perpendicular to a uniform
magnetic field is given by
View answer and explanation
Correct option: D
Fleming's left hand rule
Option D is correct. Fleming's left-hand rule gives the direction of the force on a current-carrying conductor in a magnetic field.
58 Easy
58 of 280
Application +4 | -1
10 A current is passing through a very long wire of radius 5 cm5 \mathrm{~cm}. Then magnetic field at a distance of 2 cm\mathbf{2} \mathbf{~ c m} inside from its curved surface is_______\_\_\_\_\_\_\_ ×10−5 T\times 10^{-5} \mathrm{~T}
View answer and explanation
Correct option: A
2.4×10−52.4 \times 10^{-5}
Option A is correct because the magnetic field inside the wire is calculated using Ampere's law, considering the current enclosed by the Amperian loop.
59 Medium
59 of 280
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
Option C (Zero) is correct because the net magnetic force on a closed current loop in a uniform magnetic field is zero.
60 Medium
60 of 280
Understanding +4 | -1
Force on current carrying loop (Radius =R=R) in uniform magnetic (B)(B) field which is at an angle 30∘30^{\circ} with the normal will be:
View answer and explanation
Correct option: A
0
Option A is correct. The net force on a current-carrying loop in a uniform magnetic field is zero, regardless of the angle.
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 31-60 of 280 questions

Your practice

0 / 280 answered · 280 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.