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Electromagnetic Induction MCQs for NEET — Physics Questions with Answers

Practice free Electromagnetic Induction (Physics) NEET multiple-choice questions online with instant answers and detailed explanations. No login required.

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A coil wire of a certain radius has 600 turns and a self-inductance of 36 mH. The self-inductance of a second similar coil of 500 turns will be

25 mH

50mH

108mH

75mH

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Explanation

$(1) \frac{L_{1}}{L_{2}} = \frac{N_{1}^{2}}{N_{2}^{2}} \frac{36}{L_{2}} = {(\frac{600}{500})}^{2} L_{2} = 25 mH$

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The current passing through a choke coil of 5 H is decreasing at the rate of 3 As^-1. The emf induced in the coil is

15 V

-15 V

$\frac{5}{3} V$

$\frac{- 5}{3} V$

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Explanation

$e = - L \frac{dl}{dt} = 5 \times 3 = 15 volt (c u r r e n t i s d e c r e a s d i n g)$

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A coil of resistance 20 $Ω$ and inductance 5H has been connected to a 200 V battery. The maximum energy stored in the coil is

250 J

125 J

500 J

100 J

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Explanation

$U = \frac{1}{2} L \frac{E^{2}}{R^{2}} U = \frac{1}{2} \times 5 \times 100 = 250 J$

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A coil has, 1,000 turns and 500 $c m^{2}$ as its area. The plane of the coil is placed at right angles to a magnetic induction field of $2 \times 10^{- 5} W b / m^{2}$ . The coil is rotated through 180 $°$ in 0.2. The average e.m.f. induced in the coil, in milli-volts, is

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Explanation

$e - \frac{d ϕ}{d t} = \frac{2 n B A}{t} e = \frac{2 \times 1000 \times 2 \times 10^{- 5} \times 500 \times 10^{- 4}}{0.2} e = \frac{4 \times 5 \times 10^{- 4}}{2 \times 10^{- 1}} = \frac{20}{2} \times 10^{- 3} = 10 m v$

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A small square loop of wire of side l is placed inside a large square of wire of side L (L>>l). The loops are co-planar and their centres coincide. The mutual inductance of the system is proportional to :

l/L

$l^{2} / L$

L/l

$L^{2} / l$

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Explanation

$ϕ = B A = \frac{4 \times μ i (2 \sin θ)}{4 π L} \times l^{2} M u t u a l i n d u c \tan c e M \propto \frac{l^{2}}{L}$

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Two coils have a mutual inductance of 5 mH. Current changes in the first coil according to the equation I = $I_{0}$ cos wt, where $I_{0}$ = 10 A and $ω = 100 π$ rad/s. Maximum value of e.m.f. induced in the second coil is

5 $π$ volt

2 $π$ volt

4 $π$ volt

$π$ volt

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Explanation

$(1) ε = - M \frac{dl}{dt} ε = - M \frac{d}{dt} (I_{0} \cos ω t) ε = + {MI}_{0} ωsin ωt e = + {MI}_{0} w \sin wt ε_{max} = {Ml}_{0} ω = 5 \times 10^{- 3} \times 10 \times 100 π ε_{max} = {Ml}_{0} ω = 5 \times 10^{- 3} \times 10 \times 100 π = 5 π$

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Lenz's law is a consequence of the law of conservation of

Charge

Momentum

Mass

Energy

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Explanation

The energy of the field increases with the magnitude of the field. Lenz’s law infers that there is an opposite field created due to increase or decrease of magnetic flux around a conductor so as to hold the law of conservation of energy.

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The magnetic flux through a circuit of resistance R changes by an amount $Δ φ$ in time $Δ t$ , Then the total quantity of electric charge Q, which passing during this time through any point of the circuit is given by

$Q = \frac{Δ ϕ}{Δ t}$

$Q = \frac{Δ ϕ}{Δ t} \times R$

$Q = - \frac{Δ ϕ}{Δ t} + R$

$Q = \frac{Δ ϕ}{R}$

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Explanation

We know that $e = \frac{d ϕ}{d t}$

But e=iR and $i = \frac{d q}{d t} \Rightarrow \frac{d q}{d t} R = \frac{d ϕ}{d t}$ ⇒ $d q = \frac{d ϕ}{R}$

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A coil having an area A₀ is placed in a magnetic field which changes from B₀to 4B₀ in a time interval t. The e.m.f. induced in the coil will be

$\frac{3 A_{0} B_{0}}{t}$

$\frac{4 A_{0} B_{0}}{t}$

$\frac{3 B_{0}}{A_{0} t}$

$\frac{4 B_{0}}{A_{0} t}$

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Explanation

$e = - \frac{d ϕ}{d t} = \frac{- 3 B_{0} A_{0}}{t}$

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A copper ring is held horizontally and a bar magnet is dropped through the ring with its length along the axis of the ring. The acceleration of the falling magnet while it is passing through the ring is

Equal to that due to gravity

Less than that due to gravity

More than that due to gravity

Depends on the diameter of the ring and the length of the magnet

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