A wheel with 10 metallic spokes each 1.0 m long is rotated with a speed of 240 rev/min in a plane normal to magnetic field of 0.4 G at the place, the induced emf between the axle and the rim of the wheel is;

Given that $1G=10^{-4} T$.

$5×10^{-4} V$

The correct answer is Option (4) → $5×10^{-4} V$

Given:

Number of spokes = 10 (not affecting emf)

Length of each spoke, $r = 1.0\,m$

Magnetic field, $B = 0.4\,G = 0.4 \times 10^{-4}\,T = 4 \times 10^{-5}\,T$

Speed = 240 rev/min = $\frac{240}{60} = 4\,rev/s$

Angular speed, $\omega = 2\pi f = 2\pi \times 4 = 8\pi\,rad/s$

Induced emf between axle and rim of rotating wheel:

$e = \frac{1}{2} B \omega r^2$

Substituting values:

$e = \frac{1}{2} \times 4 \times 10^{-5} \times 8\pi \times (1)^2$

$e = 2 \times 10^{-5} \times 8\pi$

$e = 16\pi \times 10^{-5}$

$e = 5.03 \times 10^{-4}\,V$

Final Answer:

$e = 5.0 \times 10^{-4}\,V$