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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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The inductance of a coil is 60μH. A current in this coil increases from 1.0 A to 1.5 A in 0.1 second. The magnitude of the induced e.m.f. is

60 × 10^–6 V

300 × 10^–4 V

30 × 10^–4 V

3 × 10^–4 V

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Explanation

$e = L \frac{d i}{d t} = 60 \times 10^{- 6} . \frac{(1 .5 - 1 .0)}{0 .1} = 3 \times 10^{- 4} v o l t$

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The self inductance of a coil is L. Keeping the length and area same, the number of turns in the coil is increased to four times. The self inductance of the coil will now be

$\frac{1}{4} L$

4 L

16 L

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Explanation

$L = \frac{μ_{0} N^{2} A}{l} L \propto N^{2} L_{f} = {(\frac{4 N}{N})}^{2} L = 16 L$

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A coil has an inductance of 2.5 H and a resistance of 0.5 r. If the coil is suddenly connected across a 6.0 volt battery, then the time required for the current to rise 0.63 of its final value is

3.5 sec

4.0 sec

4.5 sec

5.0 sec

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Explanation

$t = τ = \frac{L}{R} = \frac{2 .5}{0 .5} = 5 \sec$

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If a current of 10 A flows in one second through a coil, and the induced e.m.f. is 10 V, then the self-inductance of the coil is

$\frac{2}{5} H$

$\frac{4}{5} H$

$\frac{5}{4} H$

1 H

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Explanation

$| e | = L \frac{d i}{d t} \Rightarrow 10 = L \times \frac{10}{1} \Rightarrow L = 1 H$

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An inductance L and a resistance R are first connected to a battery. After some time the battery is disconnected but L and R remain connected in a closed circuit. Then the current reduces to 37% of its initial value in

RL sec

$\frac{R}{L} s e c$

$\frac{L}{R} s e c$

$\frac{1}{L R} s e c$

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Explanation

When battery disconnected current through the circuit start decreasing exponentially according to $i = i_{0} e^{- R t / L}$

⇒ $0 .37 i_{0} = i_{0} e^{- R t / L}$

⇒ $0 .37 = \frac{1}{e} = e^{- R t / L} \Rightarrow e = e^{\frac{L}{R t}} \Rightarrow \frac{L}{R t} = 1 \Rightarrow \frac{L}{R} = t$

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In an LR-circuit, the time constant is that time in which current grows from zero to the value (where I₀ is the steady-state current)

0.63 I₀

0.50 I₀

0.37 I₀

I₀

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Explanation

Current at any instant of time t after closing an L-R circuit is given by $I = I_{0} [1 - e^{\frac{- R}{L} t}]$

Time constant $t = \frac{L}{R}$

$∴ I = I_{0} [1 - e^{\frac{- R}{L} \times \frac{L}{R}}] = I_{0} (1 - e^{- 1}) = I_{0} (1 - \frac{1}{e})$

$= I_{0} (1 - \frac{1}{2 .718}) = 0 .63 I_{0} = 63 %$ of I₀

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