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Question
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.
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Solution
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
Current flows for time, t = 10−3 s
Average back emf, `"e" = ("d"phi)/("dt")` .........(1)
Where,
`"d"phi` = Change in flux = NAB …........(2)
Where,
B = Magnetic field strength = `mu_0("NI")/"l"` ......(3)
Where,
`mu_0` = Permeability of free space = 4π × 10−7 T m A−1
Using equations (2) and (3) in equation (1), we get
e = `(mu_0"N"^2"IA")/("lt")`
= `(4pi xx 10^-7 xx (500)^2 xx 2.5 xx 25 xx 10^-4)/(0.3 xx 10^-3)`
= 6.5 V
Hence, the average back emf induced in the solenoid is 6.5 V.
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