Solveeit Logo
Login

Question

Physics Question on Electromagnetic induction

A wheel with 2020 metallic spokes each of length 0.8 m0.8 \,m long is rotated with a speed of 120120 revolution per minute in a plane normal to the horizontal component of earth magnetic field HH at a place. If H=0.4×10−4 TH = 0.4 \times 10^{-4}\, T at the place, then induced emf between the axle and the rim of the wheel is

A

2.3×10−4 V2.3\times 10^{-4}\,V

B

3.1×10−4 V3.1\times 10^{-4}\,V

C

2.9×10−4 V2.9\times 10^{-4}\,V

D

1.61×10−4 V1.61\times 10^{-4}\,V

Answer

1.61×10−4 V1.61\times 10^{-4}\,V

Explanation

Solution

Here, H=B=0.4×10−4TH = B = 0.4\times10^{-4}T, l=0.8 ml=0.8\,m υ=120rpm=2 rps\upsilon =120 rpm =2\, rps Emf induced across the ends of each spoke ε=12Bωl2=12B(2πυ)l2(∵ω=2πυ)\varepsilon=\frac{1}{2}B\omega l^{2}=\frac{1}{2}B \left(2\pi\upsilon\right)l^{2} \left(\because\omega=2\pi\upsilon\right) =Bπυl2=B\pi\upsilon l^{2} ∴ε=0.4×10−4×π×2×(0.8)2\therefore\varepsilon=0.4\times10^{-4}\times\pi\times2\times\left(0.8\right)^{2} =1.61×10−4 V=1.61\times10^{-4}\,V