An air-cored solenoid with length 30 cm,


An air-cored solenoid with length 30 cm, area of cross-section 25 cm2 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.


Length of the solenoid, l = 30 cm = 0.3 m

Area of cross-section, A = 25 cm2 = 25 × 10−4 m2

Number of turns on the solenoid, N = 500

Current in the solenoid, I = 2.5 A

Average back emf, $e=\frac{d \phi}{d t}$    ...(1)


$d \phi=$ Change in flux

NAB … (2)


B = Magnetic field strength

$=\mu_{0} \frac{N I}{l}$   ..(3)


$\mu_{0}=$ Permeability of free space $=4 \pi \times 10^{-7} \mathrm{~T} \mathrm{~m} \mathrm{~A}^{-1}$

Using equations (2) and (3) in equation (1), we get

$e=\frac{\mu_{0} N^{2} I A}{l t}$

$=\frac{4 \pi \times 10^{-7} \times(500)^{2} \times 2.5 \times 25 \times 10^{-4}}{0.3 \times 10^{-3}}=6.5 \mathrm{~V}$

Hence, the average back emf induced in the solenoid is 6.5 V.


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