The magnetic field in a coil of 100 turns and | vedclass

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(C) According to Faraday's law of electromagnetic induction,the induced $e.m.f.$ is given by $e = -N \frac{d\phi}{dt}$.Since the magnetic field $B$ is

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(C) According to Faraday's law of electromagnetic induction,the induced $e.m.f.$ is given by $e = -N \frac{d\phi}{dt}$.Since the magnetic field $B$ is perpendicular to the coil,the angle $	heta$ between the area vector and the magnetic field is $0^\circ$,so $\cos 	heta = 1$.The magnetic flux $\phi = B \cdot A \cdot \cos 	heta = B \cdot A$.Given: $N = 100$,$A = 40 	ext{ cm}^2 = 40 	imes 10^{-4} 	ext{ m}^2$,$\Delta B = (6 - 1) 	ext{ T} = 5 	ext{ T}$,and $\Delta t = 2 	ext{ s}$.Substituting the values: $|e| = N \cdot A \cdot \frac{\Delta B}{\Delta t} = 100 	imes (40 	imes 10^{-4}) 	imes \frac{5}{2}$.$|e| = 100 	imes 0.0040 	imes 2.5 = 0.4 	imes 2.5 = 1 	ext{ V}$.

The magnetic field in a coil of $100$ turns and $40 \text{ cm}^2$ area is increased from $1 \text{ T}$ to $6 \text{ T}$ in $2 \text{ s}$. The magnetic field is perpendicular to the coil. The $e.m.f.$ generated in it is $...... \text{ V}$.

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