A circular coil of radius 10 ; cm, 500 turns, | vedclass

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Question

(C) Given: Radius $r = 0.1 \; m$, Number of turns $N = 500$, Resistance $R = 2 \; \Omega$, Time $\Delta t = 0.25 \; s$, Magnetic field $B_H = 3.0 	im

Vedclass · Accepted Answer

$3.8 	imes 10^{-3} \; V$

Vedclass · Answer

(C) Given: Radius $r = 0.1 \; m$, Number of turns $N = 500$, Resistance $R = 2 \; \Omega$, Time $\Delta t = 0.25 \; s$, Magnetic field $B_H = 3.0 	imes 10^{-5} \; T$.The magnetic flux through the coil initially is $\phi_i = N B_H A \cos(0^{\circ}) = N B_H A$.After rotating by $180^{\circ}$, the flux is $\phi_f = N B_H A \cos(180^{\circ}) = -N B_H A$.The change in flux is $\Delta \phi = \phi_f - \phi_i = -2 N B_H A$.According to Faraday's law, the induced e.m.f. is $\varepsilon = -\frac{\Delta \phi}{\Delta t} = \frac{2 N B_H A}{\Delta t}$.Area $A = \pi r^2 = \pi (0.1)^2 = 0.01 \pi \; m^2$.Substituting the values: $\varepsilon = \frac{2 	imes 500 	imes (3.0 	imes 10^{-5}) 	imes (0.01 \pi)}{0.25}$.$\varepsilon = \frac{1000 	imes 3.0 	imes 10^{-5} 	imes 0.0314}{0.25} = \frac{0.03 	imes 0.0314}{0.25} = \frac{0.000942}{0.25} = 0.003768 \; V \approx 3.8 	imes 10^{-3} \; V$.

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