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NCERT Class XII Chapter
Electromagnetic Induction
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Question : 15 of 17
Marks: +1, -0
An air cored solenoid with length 30 cm, area of cross-section 25 cm2\mathrm{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 10−310^{-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.
Solution:  
Magnetic field inside solenoid
B = μ0NIl\frac{\mu_0 N I}{l}
Flux linked with solenoid
ϕi\phi_i = BAN = μ0N2AIl\frac{\mu_0 N^2 A I}{l} ... (1)
Initial flux,
ϕi\phi_i =
4π×10−7×(500)2×25×10−4×2.530×10−2\frac{4\pi \times 10^{-7} \times (500)^2 \times 25 \times 10^{-4} \times 2.5}{30 \times 10^{-2}}
Wb
ϕi\phi_i = 6.54 × 10−310^{-3} Wb
Final flux, ϕf\phi_f = 0 [I = 0]
Average back emf
eave_{av} = - ϕf−ϕit\frac{\phi_f - \phi_i}{t} = - [0−6.54×10−310−3]\left[\frac{0 - 6.54 \times 10^{-3}}{10^{-3}}\right] = 6.54 V
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