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JEE • PHYSICS • ELECTRO MAGNETIC INDUCTION

Electro magnetic induction practice · Hard

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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.

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1 Hard

1 of 5

Analyzing +4 | -1

A sheet is placed on a horizontal surface in front of a strong magnetic pole. A force is needed to:

A. hold the sheet there if it is magnetic.

B. hold the sheet there if it is non-magnetic.

C. move the sheet away from the pole with uniform velocity if it is conducting.

D. move the sheet away from the pole with uniform velocity if it is both, non-conducting and non-polar.

Choose the correct statement(s) from the options given below:

View answer and explanation

Correct option: A

A and C only

Option A (A and C only) is correct because a force is needed to hold a magnetic sheet near a magnetic pole and to move a conducting sheet away from the pole due to eddy currents.

2 Hard

2 of 5

Analyzing +4 | -1

A wire of given length is bent in the form of a ring of diameter 2 a having self inductance

(L)

, then

L

will depend upon following

View answer and explanation

Correct option: A

L \propto a^{-1}

Option A is correct because the self-inductance L of a ring depends inversely on its radius (or diameter), hence

L \propto a^{-1}

3 Hard

3 of 5

Application +4 | -1

A uniformly wound solenoid coil of self-inductance

1.8 \times \,\,{10^{ - 4}}\,\mathrm{H}

and resistance

\mathrm{6}\,\,\mathrm{\Omega }

is broken up into two identical coils. These identical coils are then connected in parallel across a

\mathrm{12}\,\,\mathrm{V}

battery of negligible resistance. The time constant for the current in the circuit is

View answer and explanation

Correct option: C

0.3\,\, \times {10^{ - 4}}\,\mathrm{s}

Option C (

0.3\, \times {10^{ - 4}}\,\mathrm{s}

) is correct because it is the time constant calculated after breaking the coil and connecting them in parallel.

4 Hard

4 of 5

Application +4 | -1

A coil of '

\mathrm{n}

' turns and resistance '

\mathrm{R}

\Omega

is connected in series with a resistance

\frac{\mathrm{R}}{2}

. The combination is moved for time '

\mathrm{t}

' second through magnetic flux

\phi_1

\phi_2

. The induced current in the circuit is

View answer and explanation

Correct option: D

\frac{2 \mathrm{n}\left(\phi_1-\phi_2\right)}{3 \mathrm{Rt}}

Option D (

\frac{{2 \mathrm{n}(\phi_1-\phi_2)}}{{3 \mathrm{Rt}}}

) is correct because it is derived from calculating the induced emf and then dividing by the total resistance of the circuit.

5 Hard

5 of 5

Application +4 | -1

A circular coil of area

0.01 \mathrm{~m}^2

and 40 turns is rotated about its vertical diameter with an angular speed of

50 \mathrm{~rad~s}^{-1}

in a uniform horizontal magnetic field

0.05 \mathrm{~T}

. If the average power loss due to joule heating is

25 \mathrm{~m W}

. Then, the closed loop resistance of the coil is-

View answer and explanation

Correct option: D

20 \Omega

Option D (20 Ω) is correct because the closed loop resistance is calculated from the induced EMF and average power loss.

Revision notes and key takeaways

Review ELECTRO MAGNETIC INDUCTION in your textbook or class notes, then use this set to check your understanding.

Key Takeaways

✓ This set contains 5 single-correct-answer MCQs on ELECTRO MAGNETIC INDUCTION.
✓ Try each question before opening its answer. Read the explanation where one is available.
✓ Sign in to save submitted answers. Revisit incorrect answers alongside your textbook or class notes.

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