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Question

Physics Question on Electromagnetic induction

A circular coil of radius 6cm6\, cm and 2020 turns rotates about its vertical diameter with an angular speed of 40rads140\, rad \, s^{-1} in a uniform horizontal magnetic field of magnitude 2×102T2 \times 10^{-2}\, T. If the coil form a closed loop of resistance 8ω8 \,\omega, then the average power loss due to joule heating is

A

2.07×103W2.07 \times 10^{-3}\, W

B

1.23×103W1.23 \times 10^{-3}\, W

C

3.14×103W3.14 \times 10^{-3}\, W

D

1.80×103W1.80 \times 10^{-3}\, W

Answer

2.07×103W2.07 \times 10^{-3}\, W

Explanation

Solution

Here, r=6cm=6×102mr = 6 cm = 6 \times 10^{-2}\,m, N=20,ω=40rads1N=20, \omega=40\,rad\,s^{-1} B=2×102T,R=8ΩB=2\times10^{-2}\,T, R=8\, \Omega Maximum emf induced, ε=NABω\varepsilon=NAB\omega =N(πr2)Bω=N\left(\pi r^{2}\right)B\omega =20×π×(6×102)2×2×102×40=20\times\pi\times\left(6\times10^{-2}\right)^{2}\times2\times10^{-2}\times40 =0.18V=0.18\,V Average value of emf induced over a full cycle εav=0\varepsilon_{av}=0 Maximum value of current in the coil, I=εR=0.188=0.023AI=\frac{\varepsilon}{R}=\frac{0.18}{8}=0.023\,A Average power dissipated, p=εI2=0.18×0.0232p=\frac{\varepsilon I}{2}=\frac{0.18\times0.023}{2} =2.07×103W=2.07\times10^{-3}\,W