Question 8: Electromagnetic Induction

Question

A circular coil of area $2 \sqrt{2} \mathrm{~cm}^2$ and resistance $2 \boldsymbol{\Omega}$ is arranged vertically in the east -west direction. A uniform magnetic field 0.2 T is set up across the plane in the north to east direction. Now the magnetic field is removed at a steady rate in 0.4 s What is the current developed in the coil?

Accepted Answer

LetÃ¢ÂÂs go step by step:
Convert the area into SI units:
$$A = 2\sqrt2\;	ext{cm}^2 = 2\sqrt2	imes10^{-4}\;\mathrm m^2.$$
Orientation Ã¢ÂÂ only the component of
B
perpendicular to the coilÃ¢ÂÂs plane contributes to the flux.
The coilÃ¢ÂÂs plane is vertical, its normal is along northÃ¢ÂÂsouth.
The field is in the horizontal NÃ¢ÂÂE direction (i.e. 45ÃÂ° from north), so
$$B_\perp = B\cos45^\circ = 0.2	imes\frac1{\sqrt2}.$$
Initial magnetic flux
$$\Phi = B_\perp\,A = 0.2\cdot\frac1{\sqrt2}\cdot\bigl(2\sqrt2	imes10^{-4}\bigr)                 = 4	imes10^{-5}\;\mathrm Wb.$$
It goes to zero in $\Delta t=0.4$Ã¢ÂÂs, so the induced emf is
$$\mathcal E = \frac{\Delta\Phi}{\Delta t}                 = \frac{4	imes10^{-5}}{0.4}                 = 1	imes10^{-4}\;\mathrm V.$$
The coilÃ¢ÂÂs resistance is $R=2\;\Omega$, hence the induced current
$$I = \frac{\mathcal E}{R}          = \frac{1	imes10^{-4}}{2}          = 0.5	imes10^{-4}\;\mathrm A           = 5	imes10^{-5}\;\mathrm A.$$
Answer (Option B): Ã¢ÂÂ$0.5	imes10^{-4}$ A.

Answer

$0.5 	imes 10^{-3} A$

Answer

$0.5 	imes 10^{-4} A$

Answer

$1 	imes 10^{-4} A$

Answer

$0.5 	imes 10^{-5} A$

Electromagnetic Induction

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