A coil of effective area 3 text{ m}^2 is plac | vedclass

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(B) Given:
Area $A = 3 	ext{ m}^2$
Initial magnetic field $B_1 = 0.05 	ext{ Wb/m}^2$
Final magnetic field $B_2 = 20\% 	ext{ of } B_1 = 0.20 	imes

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$12$

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(B) Given:
Area $A = 3 	ext{ m}^2$
Initial magnetic field $B_1 = 0.05 	ext{ Wb/m}^2$
Final magnetic field $B_2 = 20\% 	ext{ of } B_1 = 0.20 	imes 0.05 = 0.01 	ext{ Wb/m}^2$
Time interval $\Delta t = 10 	ext{ s}$
Since the coil is placed at right angles to the field, the angle between the area vector and the magnetic field is $0^\circ$, so $\cos 	heta = 1$.
The induced e.m.f. is given by Faraday's law: $\varepsilon = -\frac{d\Phi}{dt} = -A \frac{(B_2 - B_1)}{\Delta t}$.
Substituting the values:
$\varepsilon = -3 	imes \frac{(0.01 - 0.05)}{10}$
$\varepsilon = -3 	imes \frac{(-0.04)}{10}$
$\varepsilon = \frac{0.12}{10} = 0.012 	ext{ V}$
Converting to millivolts: $0.012 	ext{ V} = 12 	ext{ mV}$.
Thus, the induced e.m.f. is $12 	ext{ mV}$.

$A$ coil of effective area $3 \text{ m}^2$ is placed at right angles to a magnetic field of induction $0.05 \text{ Wb/m}^2$. If the field is decreased to $20\%$ of its original value in $10 \text{ seconds}$, the e.m.f. induced in the coil will be (in $\text{ mV}$)

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