A flat coil, C of n turn, area A and resistance R, is placed in a uniform magnetic field of magnitude B. The plane of the coil is initially perpendicular to B. The coil is rotated by an angle θ about a diameter and charge of amount Q flows through it. Choose the correct alternatives. This question has multiple correct options A θ=90 o ,Q=(BAn/R) B θ=180 o ,Q=(2BAn/R) C θ=180 o ,Q=0 D θ=360 o ,Q=0

A flat coil, $C$ of $n$ turn, area $A$ and resistance $R$ , is placed in a uniform magnetic field of magnitude $B$ . The plane of the coil is initially perpendicular to $B$ . The coil is rotated by an angle $θ$ about a diameter and charge of amount $Q$ flows through it. Choose the correct alternatives.

This question has multiple correct options

$θ = 90^{o}, Q = (B A n / R)$

$θ = 180^{o}, Q = (2 B A n / R)$

$θ = 180^{o}, Q = 0$

$θ = 360^{o}, Q = 0$

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Solution

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Correct options are A) , B) and D)

Current through a coil rotating with angular velocity $ω$ is given by:

$i = \frac{n B A ω s i n ω t}{R}$

Charge flown on rotating the coil through an angle $θ$ is given by:

$Q = \int i d t = \int \frac{i}{ω} d (ω t) = \int \frac{i}{ω} d θ$

$Q = \frac{- n B A c o s θ}{R} + \frac{n B A}{R}$

For $θ = 9 0^{0}$ , $Q = \frac{- n B A c o s 9 0 ^{0}}{R} + \frac{n B A}{R} = 0 + \frac{n B A}{R} = \frac{n B A}{R}$

For $θ = 18 0^{0}$ , $Q = \frac{- n B A c o s 1 8 0 ^{0}}{R} + \frac{n B A}{R} = \frac{n B A}{R} + \frac{n B A}{R} = \frac{2 n B A}{R}$

For $θ = 36 0^{0}$ , $Q = \frac{- n B A c o s 3 6 0 ^{0}}{R} + \frac{n B A}{R} = \frac{- n B A}{R} + \frac{n B A}{R} = 0$

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A flat coil, C of n turn, area A and resistance R, is placed in a uniform magnetic field of magnitude B. The plane of the coil is initially perpendicular to B. The coil is rotated by an angle θ about a diameter and charge of amount Q flows through it. Choose the correct alternatives.

θ=90o,Q=(BAn/R)

θ=180o,Q=(2BAn/R)

θ=180o,Q=0

θ=360o,Q=0

$θ = 90^{o}, Q = (B A n / R)$

$θ = 180^{o}, Q = (2 B A n / R)$

$θ = 180^{o}, Q = 0$

$θ = 360^{o}, Q = 0$