First slide

Faraday's law of Electromagnetic induction

Question

A small coil of 10 turns is placed inside a solenoid of length 20 cm and 240 turns carry a current of $\frac{10}{π}$ A The area of small coil is 2.5 cm² and resistance 4.8 $Ω$ then the current reduces to zero in 25 ms, the value of average induced current is :

Easy

A

1 mA
B

2 mA
C

0.1 mA
D

0.4 mA

Solution

Values given for coil :
Number of turns of small coil = 10
Area of coil= 2.5 cm²= 2.5 x 10^-4m²
$∆ t = 25 x 10^{- 3} s$
Values given for solenoid
I = 20 x 10^-2m, N = 240
$I = \frac{10}{π} amp, n = \frac{240}{20 x 10^{- 2}} = 12 x 10^{2}$
According to Faraday's law, induced emf $E = - \frac{dϕ}{dt}$
Induced emf in coil :
$I = \frac{|- \frac{dϕ}{dt}|}{R} = \frac{(NBA) / dt}{R} = \frac{μ_{0} nI (NA)}{∆ tR} = \frac{10 x 4 π x 10^{- 7} x 12 x 10^{2} x 10 / π x 2.5 x 10^{- 4}}{25 x 10^{- 3} x 4.8} = \frac{4 x 12 x 10^{- 7} x 10^{2} x 10 x 10^{- 4}}{10^{- 3} x 4.8} = 10^{- 4} I = 0.1 mA$

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