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A magnetic field $B=2 t+4 t^{2}$ (where, $t_{x}$ time) is applied perpendicular to the plany of a circular wire of radius $r$ and resistance
R. If all the units are in SI the electric charge that flows through the circular wire during $t=0 \mathrm{s}$ to $t=2 \mathrm{s}$ is
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R. If all the units are in SI the electric charge that flows through the circular wire during $t=0 \mathrm{s}$ to $t=2 \mathrm{s}$ is
Solution:
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Verified Answer
The correct answer is:
$\frac{20 \pi r^{2}}{R}$
Given, $B=2 t+4 t^{2}$
at $\quad t=0, B_{1}=0$
and at $t=2, B_{2}=2 \times 2+4(2)^{2}$
$$
=4+16=20 \mathrm{Wb} / \mathrm{m}^{2}
$$
We have, $\Delta Q=\frac{\Delta \phi}{R}=\frac{\pi^{2}\left(B_{2}-B_{1}\right)}{R}$
$$
=\frac{\pi r^{2} \left[20-0\right]}{R}=\frac{20 \pi r^{2}}{R}
$$
at $\quad t=0, B_{1}=0$
and at $t=2, B_{2}=2 \times 2+4(2)^{2}$
$$
=4+16=20 \mathrm{Wb} / \mathrm{m}^{2}
$$
We have, $\Delta Q=\frac{\Delta \phi}{R}=\frac{\pi^{2}\left(B_{2}-B_{1}\right)}{R}$
$$
=\frac{\pi r^{2} \left[20-0\right]}{R}=\frac{20 \pi r^{2}}{R}
$$
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