A conducting ring of radius $1 m$ is placed in a uniform magnetic field of $0.01 T$ oscillating with frequency $100 H z$ with its plane at right angle to $B$ . What will be the induced electric field?

π V / m

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2 V / m

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10 V m^{- 1}

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62 V m^{- 1}

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Solution

The correct option is D $2 V / m$
After every $0.5 T$ the field will change from $B$ to $- B$
$ϵ = \frac{2 B A}{T / 2} = 4 B A f$
$\oint E . d l = ϵ$
or $E = \frac{ϵ}{2 π R}$
$= \frac{4 B π R^{2} f}{2 π R}$
$= B R f = 2 \times 0.01 \times 1 \times 100 = 2 V m^{- 1}$

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A conducting ring of radius 1m is placed in a uniform magnetic field of 0.01T oscillating with frequency 100Hz with its plane at right angle to B. What will be the induced electric field?

The correct option is D 2V/mAfter every 0.5T the field will change from B to −B ϵ=2BAT/2=4BAf∮E.dl=ϵor E=ϵ2πR=4BπR2f2πR=BRf=2×0.01×1×100=2Vm−1

A conducting ring of radius $1 m$ is placed in a uniform magnetic field of $0.01 T$ oscillating with frequency $100 H z$ with its plane at right angle to $B$ . What will be the induced electric field?

The correct option is D $2 V / m$
After every $0.5 T$ the field will change from $B$ to $- B$
$ϵ = \frac{2 B A}{T / 2} = 4 B A f$
$\oint E . d l = ϵ$
or $E = \frac{ϵ}{2 π R}$
$= \frac{4 B π R^{2} f}{2 π R}$
$= B R f = 2 \times 0.01 \times 1 \times 100 = 2 V m^{- 1}$