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Question 6.3: A long solenoid with 15 turns per cm has a small loop of area 2.0 cm 2 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?Class 12 - Physics - Electromagnetic Induction Electromagnetic Induction Page-230

Answer»

Number of turns on the solenoid = 15 turns/cm = 1500 turns/m
so, number of turns per unit length, N = 1500
the solenoid has a small loop of area , A = 2 cm² = 2 × 10^-4 cm²
CURRENT carries by the solenoid change from 2A to 4A .
so, change in current in the solenoid , ∆i = 4 - 2 = 2A
change in time , ∆t = 0.1 s

now, induced emf , e=\frac{d\phi}{dt}
where \phi is the induced flix through the small loop , e.g., BA [ flux = magnetic field× area]
so,e=\frac{d}{dt}(BA)
for solenoid B = \mu_0 ni
so, e=\mu_0 nA\frac{di}{dt}
= 4π × 10^-7 × 1500 × 2 × 10^-4 × (2/0.1)
= 7.54 × 10^-6 V

HENCE, the induced VOLTAGE in the loop is 7.54 × 10^-6V


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