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

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(A) The magnetic field inside an air-cored solenoid is given by $B = \frac{\mu_0 NI}{l}$.Initial magnetic flux linkage $\phi_1 = N B A = N \left( \fra

Vedclass · Accepted Answer

$6.54$

Vedclass · Answer

(A) The magnetic field inside an air-cored solenoid is given by $B = \frac{\mu_0 NI}{l}$.Initial magnetic flux linkage $\phi_1 = N B A = N \left( \frac{\mu_0 NI}{l} ight) A = \frac{\mu_0 N^2 AI}{l}$.Given values: $N = 500$,$I = 2.5 \ A$,$l = 0.3 \ m$,$A = 25 	imes 10^{-4} \ m^2$,$\Delta t = 10^{-3} \ s$,and $\mu_0 = 4\pi 	imes 10^{-7} \ T \cdot m/A$.When the current is switched off,the final flux $\phi_2 = 0$.The magnitude of induced emf is $|\varepsilon| = \frac{|\Delta \phi|}{\Delta t} = \frac{|\phi_2 - \phi_1|}{\Delta t} = \frac{\mu_0 N^2 AI}{l \Delta t}$.Substituting the values:$|\varepsilon| = \frac{4\pi 	imes 10^{-7} 	imes (500)^2 	imes 25 	imes 10^{-4} 	imes 2.5}{0.3 	imes 10^{-3}}$.$|\varepsilon| = \frac{4 	imes 3.1416 	imes 10^{-7} 	imes 250000 	imes 25 	imes 10^{-4} 	imes 2.5}{3 	imes 10^{-4}}$.$|\varepsilon| = \frac{1.9635 	imes 10^{-3}}{3 	imes 10^{-4}} \approx 6.54 \ V$.

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