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Question 6.15: An air-cored solenoid with length 30 cm, area of cross-section 25 cm 2 and number of turns 500, carries a current of 2.5 A. The current is suddenly switched off in a brief time of 10 −3 s. How much is the average back emf induced across the ends of the open switch in the circuit? Ignore the variation in magnetic field near the ends of the solenoid.Class 12 - Physics - Electromagnetic Induction Electromagnetic Induction Page-232

Answer» MAGNETIC FIELD INSIDE solenoid is B=\frac{\mu_0NI}{l}
flux linked with solenoid is \phi=BAN=\frac{\mu_0N^2AI}{l}
so, the initial flux is \phi_i=\frac{\mu_0N^2AI}{l}
\phi_i=\frac{4\pi\times500^2\times25\times10^{-4}\times2.5}{30\times10^{-2}}Wb\\\\\phi_i=6.54\times10^{-3}Wb
final flux , \phi_f=0 [because current I=0]
AVERAGE back emf ,
e_{av}=-\frac{\phi_f-\phi_i}{t}=-\frac{0-6.54\times10^{-3}}{10^{-3}}V\\\\e_{av}=6.54V


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