First slide

Magnetic flux and induced emf

Question

Shown in the figure is a circular loop of radius r and resistance R. A variable magnetic field of induction B = B₀e^-t is established inside the coil. If the key (K) is closed, the electrical power developed right after closing the switch is equal to

Difficult

A

$\frac{B_{0}^{2} π r^{2}}{R}$
B

$\frac{B_{0} 10 r^{3}}{R}$
C

$\frac{B_{0}^{2} π^{2} r^{4} R}{5}$
D

$\frac{B_{0}^{2} π^{2} r^{4}}{R}$

Solution

$\begin{array}{l} P = \frac{ε^{2}}{R}; ε = - \frac{d}{d t} (B A) = - A \frac{d}{d t} (B_{o} e^{- t}) = A B_{o} e^{- t} \\ \Rightarrow P = \frac{1}{R} {(A B_{o} e^{- t})}^{2} = \frac{A^{2} B_{o}^{2} e^{- 2 t}}{R} \end{array}$
At the time of starting $t = 0 so P = \frac{A^{2} B_{o}^{2}}{R}$
$\Rightarrow P = \frac{{(π r^{2})}^{2} B_{o}^{2}}{R} = \frac{B_{o}^{2} π^{2} r^{4}}{R}$

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