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Standard XII

Physics

Magnetic Flux and Faraday's Law

A flexible wi...

Question

A flexible wire loop in the shape of a circle has a radius that grows linearly with time. There is a magnetic field perpendicular to the plane of the loop that has a magnitude inversely proportional to the distance from the center of the loop, $B (r) \propto \frac{1}{r}$ . How does the emf vary with time?

E \propto r^{2}

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E \propto t

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E \propto \sqrt{t}

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E

is constant

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Solution

The correct option is D $E$ is constant
Given
Radius = $r = k t$
.
Also $B (r) = \frac{a}{r}$
Magnetic flux = $ϕ = \int B . d A = \int_{0}^{r} \frac{a}{r} 2 π r d r = 2 π r = 2 a π k t$
By faraday's law
$E = - \frac{d ϕ}{d t} = 2 a π k$

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A flexible wire loop in the shape of a circle has a radius that grows linearly with time. There is a magnetic field perpendicular to the plane of the loop that has a magnitude inversely proportional to the distance from the center of the loop, B(r)∝1r. How does the emf vary with time?

The correct option is D E is constantGivenRadius = r=kt.Also B(r)=arMagnetic flux = ϕ=∫B.dA=∫r0ar2πrdr=2πr=2aπktBy faraday's lawE=−dϕdt=2aπk

The correct option is D $E$ is constant
Given
Radius = $r = k t$
.
Also $B (r) = \frac{a}{r}$
Magnetic flux = $ϕ = \int B . d A = \int_{0}^{r} \frac{a}{r} 2 π r d r = 2 π r = 2 a π k t$
By faraday's law
$E = - \frac{d ϕ}{d t} = 2 a π k$