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

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(B) Length of the solenoid,$l = 30 \; cm = 0.3 \; m$Area of cross-section,$A = 25 \; cm^{2} = 25 	imes 10^{-4} \; m^{2}$Number of turns on the soleno

Vedclass · Accepted Answer

$6.5$

Vedclass · Answer

(B) Length of the solenoid,$l = 30 \; cm = 0.3 \; m$Area of cross-section,$A = 25 \; cm^{2} = 25 	imes 10^{-4} \; m^{2}$Number of turns on the solenoid,$N = 500$Current in the solenoid,$I = 2.5 \; A$Time interval,$\Delta t = 10^{-3} \; s$The magnetic field inside the solenoid is $B = \mu_{0} \frac{N I}{l}$.The magnetic flux through each turn is $\phi = B A = \frac{\mu_{0} N I A}{l}$.The total flux linkage is $N \phi = \frac{\mu_{0} N^{2} I A}{l}$.The average induced back $emf$ is given by $e = \frac{\Delta(N \phi)}{\Delta t} = \frac{\mu_{0} N^{2} I A}{l \Delta t}$.Substituting the values:$e = \frac{4 \pi 	imes 10^{-7} 	imes (500)^{2} 	imes 2.5 	imes 25 	imes 10^{-4}}{0.3 	imes 10^{-3}}$$e = \frac{4 	imes 3.14159 	imes 10^{-7} 	imes 250000 	imes 2.5 	imes 25 	imes 10^{-4}}{0.3 	imes 10^{-3}}$$e = \frac{3.14159 	imes 10^{-7} 	imes 10^{6} 	imes 62.5 	imes 25 	imes 10^{-4}}{0.3 	imes 10^{-3}}$$e \approx 6.54 \; V \approx 6.5 \; V$.Thus,the average back $emf$ induced is $6.5 \; V$.

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