Question

A wheel with 20 metallic spokes each of length 0.8 m long is rotated with a speed of 120 revolution per minute in a plane normal to the horizontal component of earth magnetic field H at a place. If H=$0.4\times {10}^{-4}$ T at the place, then induced emf between the axle and the rim of the wheel is

• A. $2.3\times {10}^{-4}$
• B. $3.1\times {10}^{-4}$
• C. $2.9\times {10}^{-4}$
• D. $1.61\times {10}^{-4}$

Answer 1

Answer: (D)

$1.61\times {10}^{-4}$

Here, $H=B=0.4\times {10}^{-4}T,l=0.8mv=120rpm=2rps$

Emf induced across the ends of each spoke

$ϵ=\frac{1}{2}B\omega l^2=\frac{1}{2}B\left(2\pi \upsilon \right)l^2$ $(\because \omega =2\pi \upsilon )$

$=B\pi \upsilon l^2$

$\therefore ϵ=0.4\times {10}^{-4}\times \pi \times 2\times (0.8{)}^2=1.61\times {10}^{-4}V$

Note : Number of spokes is not relevant because the emfs across the spoke area is parallel.