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NEET UG PHYSICS MCQ: ELECTRO MAGNETIC INDUCTION

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What is the self inductance of an air core solenoid 1 m long, diameter 0.05m, if it has 500 turns? Take

{\pi ^2} = 10.

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Correct option: D

6.25\; \times {10^{ - 4}}H

Option D (

6.25\; \times {10^{ - 4}}H

) is correct because it's the self-inductance calculated using the formula for a solenoid with the given parameters.

More Questions from ELECTRO MAGNETIC INDUCTION

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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 