An air-cored solenoid with length 30 cm,area | vedclass

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(B) The magnetic field $B$ at the center of the solenoid is given by $B = \mu_0 (N/l) i$.The magnetic flux $\Phi$ linked with the solenoid is $\Phi =

Vedclass · Accepted Answer

$6.5$

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(B) The magnetic field $B$ at the center of the solenoid is given by $B = \mu_0 (N/l) i$.The magnetic flux $\Phi$ linked with the solenoid is $\Phi = N B A = N (\mu_0 N i A / l) = (\mu_0 N^2 A / l) i$.The self-inductance $L$ is defined as $L = \Phi / i = \mu_0 N^2 A / l$.Substituting the given values: $N = 500$,$l = 0.3 \ m$,$A = 25 	imes 10^{-4} \ m^2$,$\mu_0 = 4\pi 	imes 10^{-7} \ T \cdot m/A$.$L = \frac{4\pi 	imes 10^{-7} 	imes (500)^2 	imes 25 	imes 10^{-4}}{0.3} \approx 2.618 	imes 10^{-3} \ H$.The average back e.m.f. is $e = L (\Delta i / \Delta t)$.Given $\Delta i = 2.5 \ A$ and $\Delta t = 10^{-3} \ s$.$e = (2.618 	imes 10^{-3}) 	imes (2.5 / 10^{-3}) = 2.618 	imes 2.5 \approx 6.545 \ V$.Rounding to the nearest value,$e \approx 6.5 \ V$.

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