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Electromagnetic Induction MCQs for NEET — Physics Questions with Answers

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An air-plane with 20m wing spread is flying at $250 {ms}^{- 1}$ straight south parallel to the earth’s surface. The earth’s magnetic field has a horizontal component of $2 \times 10^{- 5}$ $Wb m^{- 2}$ and the dip angle is 60º. Calculate the induced emf between the plane tips is:

0.174 V

0.173 V

1.173 V

0.163 V

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Explanation

2. As the plane is flying horizontally it will cut the vertical component of earth’s field $B_{V}$ . So the

emf induced between its tips, $e = B_{V} lv$

But as by definition of angle of dip,

$\tan ϕ = \frac{B_{V}}{B_{H}} i . e ., B_{V} = B_{H} \tan ϕ So e = (B_{H} \tan ϕ) lv = 2 \times 10^{- 5} \times \sqrt{3} \times 250 \times 20 i . e ., e = (\sqrt{3}) \times 10^{- 1} V = 0.173 V$

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A wire of fixed lengths is wound on a solenoid of length $l$ and radius r. Its self inductance is found to be L. Now if same wire is wound on a solenoid of length $l / 2$ and radius r/2, then the self inductance will be –

$2 L$

$L$

$4 L$

$8 L$

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Explanation

$1 . L = \frac{μ_{0} N^{2} {πr}^{2}}{l} Length of wire = N 2 πr = constant (= C, suppose) ∴ L = μ_{0} {(\frac{C}{2 πr})}^{2} \frac{{πr}^{2}}{l} ∴ L \propto \frac{1}{l} ∴ Self inductance will become 2 L .$

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A wire in the form of a circular loop of radius 10 cm lies in a plane normal to a magnetic field of 100 T. If this wire is pulled to take a square shape in the same plane in 0.1 s, average induced emf in the loop is:

6.70 volt

5.80 volt

6.75 volt

5.75 volt

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Explanation

$3 . According to Faraday ’ s law of electromagnetic induction, E_{induced} = \frac{- ∆ ϕ}{∆ t} = - \frac{B (A_{f} - A_{i})}{∆ t} Let r be the radius of circle; then side of square formed = \frac{2 πr}{4} = \frac{πr}{2} Change is area of loop = A_{i} - A_{f} = {πr}^{2} - {(\frac{πr}{2})}^{2} = \frac{π (4 - π) r^{2}}{4} Hence average emf induced = \frac{π (4 - π) r^{2}}{4} . \frac{B}{t} = \frac{π (4 - π) \times {(0.1)}^{2} \times 100}{4 \times 0.1} = 6.75 volt .$

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A superconducting loop of radius R has self inductance L. A uniform and constant magnetic field B is applied perpendicular to the plane of the loop. Initially current in this loop is zero. The loop is rotated by 180°. The current in the loop after rotation is equal to –

zero

$\frac{{BπR}^{2}}{L}$

$\frac{2 {BπR}^{2}}{L}$

$\frac{{BπR}^{2}}{2 L}$

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Explanation

3. Flux can’t change in a superconducting loop.

$∆ ϕ = 2 {πR}^{2} . B$

Initially current was zero, so self flux was zero.

$∴$ Finally $Li = 2 {πR}^{2} \times B$

$i = \frac{2 {πR}^{2} \times B}{L}$

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A coil of inductance 8.4 mH and resistance 6 $Ω$ is connected to a 12 V battery. The current in the coil is 1.0 A at approximately the time.

500 ms

20 ms

35 ms

1 ms

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Explanation

$4 . Current developed with time in a coil of inductance I = \frac{V}{R} (1 - e^{- t / τ}) where τ = L / R we have τ = \frac{8.4 mH}{6 Ω} = 1.4 ms Hence, 1 A = (\frac{12 V}{6 Ω}) (1 - e^{- t / 1.4 ms}) or e^{- t / 1.4 ms} = 1 - \frac{1}{2} = \frac{1}{2} or - t / 1.4 ms = 1 n (\frac{1}{2}) = - 0.693 or t = (1.4 \times 0.693) ms = 0.97 ms \approx 1 ms .$

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A 50 turns circular coil has a radius of 3 cms, it is kept in a magnetic field acting normal to the area of the coil. The magnetic field B increased from 0.10 tesla to 0.35 tesla in 2 milliseconds. The average induced emf in the coil is-

1.77 volts

17.7 volts

177 volts

0.177 volts

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Explanation

$2 . ϕ = NBA ϕ_{1} = 50 \times π \times {(3 \times 10^{- 2})}^{2} \times 0.1 = 141.3 \times 10^{- 4} Wb ϕ_{2} = 50 \times π \times {(3 \times 10^{- 2})}^{2} \times 0.35 = 494.5 \times 10^{- 4} Wb ∴ e = \frac{dϕ}{dt} = 17.7 volt$

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The inductance of a closed-packed coil of 400 turns is 8 mH. A current of 5 mA is passed through it. The magnetic flux through the coil is approximately

$0.1 μ_{0} Wb$

$0.2 μ_{0} Wb$

$1.0 μ_{0} Wb$

$2.0 μ_{0} Wb$

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Explanation

$1 . L = \frac{Nϕ}{i} 8 \times 10^{- 3} = \frac{400 \times ϕ}{5 \times 10^{- 3}} \Rightarrow ϕ = \frac{40 \times 10^{- 6}}{400} Wb = 10^{- 7} Wb \Rightarrow ϕ = \frac{4 π \times 10^{- 7}}{4 π} Wb \Rightarrow ϕ = \frac{μ_{0}}{4 π} Wb \Rightarrow ϕ \approx 0.1 μ_{0} Wb$

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The current in an L – R circuit builds up to ${(3 / 4)}^{th}$ of its steady state value in 4 seconds. The time constant of this circuit is

$\frac{1}{In 2} \sec$

$\frac{2}{In 2} \sec$

$\frac{3}{In 2} \sec$

$\frac{4}{In 2} \sec$

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Explanation

$2 . I = I_{0} (l - e^{- t / τ}) where τ \to time constant ∴ \frac{3}{4} I_{0} = I_{0} (l - e^{- t / τ}) \Rightarrow \frac{3}{4} = l - e^{- t / τ} \Rightarrow e^{- t / τ} = \frac{1}{4} \Rightarrow \frac{- t}{τ} In e = In \frac{1}{4} \Rightarrow \frac{- 4}{τ} = - 2 In 2 \Rightarrow τ = \frac{2}{In 2}$

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The magnetic flux through each turn of a 100 turn coil is $(t^{3} - 2 t) \times 10^{- 3} Wb$ , where t is in second. The induced emf at t = 2 s is

$- 4 V$

$- 1 V$

$+ 1 V$

$+ 4 V$

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Explanation

$2 . ϕ = (t^{3} - 2 t) \times 10^{- 3} \frac{dϕ}{dt} = (3 t^{2} - 2) \times 10^{- 3} {\frac{dϕ}{dt}}_{t = 2} = (3 \times 4 - 2) \times 10^{- 3} Wb / s = 10^{- 2} Wb / s e = - N \frac{dϕ}{dt} = - 100 \times 10^{- 2} V = - 1 V$

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An emf of 15 volt is applied in a circuit containing 5 henry inductance and 10 ohm resistance. The ratio of the currents at time t = $\infty$ and at t = 1 second is -

$\frac{e^{1 / 2}}{e^{1 / 2} - 1}$

$\frac{e^{2}}{e^{2} - 1}$

$1 - e^{- 1}$

$e^{- 1}$

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Explanation

$2 . I = I_{0} (1 - e^{- Rt / L}) I_{0} = \frac{E}{R} (Steady current) When t = \infty I_{\infty} = \frac{E}{R} (1 - e^{- \infty}) = \frac{15}{10} = 1.5 I_{1} = 1.5 (1 - e^{- R / L}) = 1.5 (1 - e^{- 2}) ∴ \frac{I_{\infty}}{I_{1}} = \frac{1}{1 - e^{- 2}} = \frac{e^{2}}{e^{2} - 1}$

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