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Question
A long solenoid with 15 turns per cm has a small loop of area 2.0 cm2 placed inside the solenoid normal to its axis. If the current carried by the solenoid changes steadily from 2.0 A to 4.0 A in 0.1 s, what is the induced emf in the loop while the current is changing?
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Solution
Number of turns on the solenoid = 15 turns/cm = 1500 turns/m
Number of turns per unit length, n = 1500 turns
The solenoid has a small loop of area, A = 2.0 cm2
= 2 × 10−4 m2
Current carried by the solenoid changes from 2 A to 4 A.
∴ Change in current in the solenoid, di = 4 − 2 = 2 A
Change in time, dt = 0.1 s
Induced emf in the solenoid is given by Faraday’s law as:
`"e" = ("d"phi)/("dt")` ........(1)
Where,
`phi` = Induced flux through the small loop = BA ...........(2)
B = Magnetic field = `mu_0"ni"` .....(3)
μ0 = Permeability of free space
= 4π × 10−7 H/m
Hence, equation (1) reduces to:
`"e" = "d"/("dt")("BA")`
= `"A"mu_0"n" xx (("di")/("dt"))`
= `2 xx 10^-4 xx 4pi xx 10^-7 xx 1500 xx 2/0.1`
= 7.54 × 10−6 V
Hence, the induced voltage in the loop is 7.54 × 10−6 V.
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