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(A) The magnetic flux $\phi$ through the coil is given by $\phi = \vec{B} \cdot \vec{S} = B \cdot \pi R^2$.Substituting $B = \frac{4}{\pi} 	imes 10^{

Vedclass · Accepted Answer

$80$

Vedclass · Answer

(A) The magnetic flux $\phi$ through the coil is given by $\phi = \vec{B} \cdot \vec{S} = B \cdot \pi R^2$.Substituting $B = \frac{4}{\pi} 	imes 10^{-3} (1 - \frac{t}{100})$ and $R = 1\, m$,we get $\phi = 4 	imes 10^{-3} (1 - \frac{t}{100})\, Wb$.The induced electromotive force $(EMF)$ is $\varepsilon = -\frac{d\phi}{dt} = -\frac{d}{dt} [4 	imes 10^{-3} (1 - \frac{t}{100})] = 4 	imes 10^{-5}\, V$.The magnetic field becomes zero when $1 - \frac{t}{100} = 0$,which gives $t = 100\, s$.The energy dissipated $E$ is given by $E = \frac{\varepsilon^2}{R_{coil}} 	imes t$.Substituting the values: $E = \frac{(4 	imes 10^{-5})^2}{2 	imes 10^{-6}} 	imes 100 = \frac{16 	imes 10^{-10}}{2 	imes 10^{-6}} 	imes 100 = 8 	imes 10^{-4} 	imes 100 = 0.08\, J$.Since $1\, J = 1000\, mJ$,$E = 0.08 	imes 1000 = 80\, mJ$.

Similar Questions

Find the current in the wire for the configuration shown in the figure. Wire $PQ$ has negligible resistance. $\vec{B}$ is the magnetic field coming out of the paper. $\theta$ is a fixed angle made by $PQ$ as it travels smoothly over two conducting parallel wires separated by a distance $d$.

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