Back to ELECTRO MAGNETIC INDUCTION

NEET UG PHYSICS MCQ: ELECTRO MAGNETIC INDUCTION

1 Hard

1 of 1

Application +4 | -1

A total charge

{Q}

is uniformly distributed over a non-conducting ring of radius

{r}

. There is a time varying magnetic field perpendicular to its plane and changing at the uniform rate of

\frac{{dB}}{{dt}}

. The magnitude of torque experienced by the ring is

View answer and explanation

Correct option: C

\frac{1}{2} {Qr}^2\left(\frac{{dB}}{{dt}}\right)

The correct answer is

\frac{1}{2} {Qr}^2\left(\frac{{dB}}{{dt}}\right)

. The torque experienced by the ring is calculated using the induced electric field and the charge distribution.

More Questions from ELECTRO MAGNETIC INDUCTION

Practice All Chapter MCQs →

Assertion (A): Whenever the magnetic flux linked with a closed coil changes there will be an induced emf as well as an induced current.

Reason (R): According to Faraday, the induced emf is inversely proportional to the rate of change of magnetic flux linked with a coil. The coefficient of mutual inductance of two coils is

6\,\,mH

. If the current flowing in one is

2\,\,ampere

, then the induced e.m.f. in the second coil will be The induced emf cannot be produced by: Consider the statements:
(I)If magnetic field,

\mathrm{B = 0}

, then magnetic flux is also zero.
(II)If magnetic flux,

\phi = 0,

then magnetic field is also zero. The coefficient of self inductance of a solenoid is 0.18 mH. If a rod of soft iron of relative permeability 900 is inserted, then the coefficient of self inductance (in mH ) will become nearly: A uniformly wound solenoidal coil of self inductance

1.8 \times {10^{ - 4}}

H and resistance 6 Ω is broken up into two identical coils. These identical coils are then connected in parallel across a 12 V battery of negligible resistance. The time constant of the current in the circuit 