A wire loop enclosing a semi-circle of radius a is located on the boundary of a uniform magnetic field of induction $B$ (figure). At the moment $t = 0$ the loop is set into rotation with a constant angular acceleration $α$ about an axis $O$ coinciding with a line of vector $\vec{B}$ on the boundary. Then the magnitude of emf induced in the loop as a function of time $t$ is

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$2 B a^{2} α t$

$\frac{B a^{2} α t}{8}$

$\frac{B a^{2} α t}{4}$

$\frac{B a^{2} α t}{2}$

answer is D.

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Detailed Solution

The flux through the small area,

$ϕ = B A = B (\frac{a θ \times a}{2}) = B \frac{a^{2}}{2} θ$

induced emf, $e = - \frac{d ϕ}{d t} = - \frac{B a^{2}}{2} (\frac{d θ}{d t}) = - \frac{B a^{2}}{2} ω$

In time $t$ , $ω = α t$

Therefore e = - $\frac{B a^{2} α t}{2}$

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