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Electromagnetic Induction MCQs for NEET — Physics Questions with Answers

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A magnet is brought towards a coil (i) speedily (ii) slowly then the induced e.m.f./induced charge will be respectively

More in first case / More in the first case

More in first case/Equal in both case

Less in first case/More in second case

Less in first case/Equal in both case

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Explanation

The magnitude of induced e.m.f. is directly proportional to the rate of change of magnetic flux. Induced charge doesn’t depend upon time.

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A coil having 500 square loops each of side 10 cm is placed normal to a magnetic flux which increases at the rate of 1.0 tesla/second. The induced e.m.f. in volts is

0.1

0.5

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Explanation

$| e | = N (\frac{Δ B}{Δ t}) . A \cos θ$

$= 500 \times 1 \times {(10 \times 10^{- 2})}^{2} \cos 0 = 5 V .$

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The magnetic field in a coil of 100 turns and 40 square cm area is increased from 1 Tesla to 6 Tesla in 2 second. The magnetic field is perpendicular to the coil. The e.m.f. generated in it is

10⁴ V

1.2 V

1.0 V

10^–2 V

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Explanation

$e = - N (\frac{Δ B}{Δ t}) . A \cos θ$

$= - 100 \times \frac{(6 - 1)}{2} \times (40 \times 10^{- 4}) \cos 0 \Rightarrow | e | = 1 V$

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The total charge induced in a conducting loop when it is moved in the magnetic field depends on

The rate of change of magnetic flux

Initial magnetic flux only

The total change in magnetic flux

Final magnetic flux only

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Explanation

$q = \frac{N}{R} d ϕ$

$∴ q \propto d ϕ$

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A coil having n turns and resistance RΩ is connected with a galvanometer of resistance 4RΩ. This combination is moved in time t seconds from a magnetic field W₁ $w e b e r / m^{2}$ to W₂ $w e b e r / m^{2}$ . The induced current in the circuit is

$- \frac{W_{2} - W_{1}}{5 R n t}$

$- \frac{n (W_{2} - W_{1})}{5 R t}$

$- \frac{(W_{2} - W_{1})}{R n t}$

$- \frac{n (W_{2} - W_{1})}{R t}$

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Explanation

$i = \frac{e}{R} = \frac{- N}{R} \frac{(ϕ_{2} - ϕ_{1})}{Δ t} = \frac{- n (W_{2} - W_{1})}{5 R t}$

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Two rails of a railway track insulated from each other and the ground are connected to a milli voltmeter. What is the reading of voltmeter, when a train travels with a speed of 180 km/hr along the track. Given that the vertical component of earth's magnetic field is 0.2 × 10^–4 weber/m² and the rails are separated by 1 metre

10^–2 volt

10^–4 volt

10^–3 volt

1 volt

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Explanation

$e = B_{v} . v l$ $= 0 .2 \times 10^{- 4} \times (\frac{180 \times 1000}{3600}) \times 1$

$= 10^{- 3} V$

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The magnitude of the earth’s magnetic field at a place is B₀ and the angle of dip is δ. A horizontal conductor of length l lying along the magnetic north-south moves eastwards with a velocity v. The emf induced across the conductor is

Zero

B₀lv sinδ

B₀lv

B₀lv cosδ

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Explanation

When a conductor lying along the magnetic north-south, moves eastwards it will cut vertical component of B₀. So induced emf

$e = v B_{V} l = v (B_{0} \sin δ l) = B_{0} l v \sin δ$

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Two coils of self inductance L₁ and L₂ are placed closer to each other so that total flux in one coil is completely linked with other. If M is mutual inductance between them, then

M = L₁ L₂

M = L₁/L₂

$M = \sqrt{L_{1} L_{2}}$

M = (L₁ L₂)²

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Explanation

$M = - \frac{e_{2}}{d i_{1} / d t} = - \frac{e_{1}}{d i_{2} / d t}$

Also $e_{1} = - L_{1} \frac{d i_{1}}{d t} a n d e_{2} = - L_{2} \frac{d i_{2}}{d t}$

$M^{2} = \frac{e_{1} e_{2}}{(\frac{d i_{1}}{d t}) (\frac{d i_{2}}{d t})} = L_{1} L_{2} \Rightarrow M = \sqrt{L_{1} L_{2}}$

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Two circuits have coefficient of mutual induction of 0.09 henry. Average e.m.f. induced in the secondary by a change of current from 0 to 20 ampere in 0.006 second in the primary will be

120 V

80 V

200 V

300 V

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Explanation

$e = M \frac{d i}{d t} = 0 .09 \times \frac{20}{0 .006} = 300 V$

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A coil and a bulb are connected in series with a dc source, a soft iron core is then inserted in the coil. Then

Intensity of the bulb remains the same

Intensity of the bulb decreases

Intensity of the bulb increases

The bulb ceases to glow

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Explanation

There will be no change in the intensity of the bulb as the reactance offered by a coil to a d.c. current is zero. So, the bulb will glow with the same intensity as earlier when the iron rod was not inserted.

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