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Question

Physics Question on Electromagnetic induction

An air cored solenoid with length 20cm20\, cm, area of cross section 20cm220\, cm^{2} and number of turns 400400 carries a current 2A2\, A. The current is suddenly switched off within 103s10^{-3}\, s. The average back emf induced across the ends of the open switch in the circuit is (ignore the variation in magnetic field near the ends of the solenoid)

A

2V2\,V

B

4V4\,V

C

3V3\,V

D

5V5\,V

Answer

4V4\,V

Explanation

Solution

Here, l=20cm=20×102ml = 20\, cm = 20 \times 10^{-2}\, m, A=20cm2=20×104m2A=20\,cm^{2}=20\times10^{-4}\,m^{2} N=400N=400, I1=2AI_{1}=2\,A, I2=0I_{2}=0, dt=103sdt=10^{-3}\,s As ε=dϕdt=d(BAN)dt\varepsilon= \frac{d \,\phi}{dt}=\frac{d(BAN)}{dt} =μ0NdlANldt(B=μ0NIt)=\frac{\mu_{0}N\,dl\,AN}{l\,dt}\left(\because B=\frac{\mu_{0}NI}{t}\right) =μ0N(I1I2)ANldt=\frac{\mu_{0}N (I_{1}-I_{2})AN}{l\,dt} =4π×107×(400)2×2×20×10420×102×103=\frac{ 4\,\pi\times10^{-7}\times(400)^{2}\times2\times20\times10^{-4}}{20\times10^{2}\times10^{-3} } =4V=4\, V