A wheel with 10 metallic spokes each 0.5 m long is rotated with a speed of 120 rev/min, in a plane normal to earth's magnetic field at the place. If the magnitude of the field is 0.40 G, what is the induced emf between the axle and rim of the wheel?

Utilize the idea of induced emf and adjust the formula to the magnetic field's direction. Rev/min to rev/sec conversion. Tesla (T) to Gauss (G) conversion. Consider a cycle as an illustration to improve your creativity.

Step 1. Given data:
The quantity of metallic spokes $\left(N\right)=10$
Wheel radius or the diameter of a spoke, $\left(r\right)$ or $\left(L\right)=0.5 \mathrm{m}$
Wheel's rotation $\left(f\right)=120\mathrm{rev}/\min =2\mathrm{rev}/\sec$
Dimensions of the field $\left(B\right)=0.4$ Gauss $=0.4\times {10}^{-4} \mathrm{T}$
Step 2. Formula used:
Between the axle and the rim is provided by,
$e=Bvl$, where each symbol is explained above.
Step 3: Calculation
We are aware that a revolution $=2\mathrm{\pi } rad$
angular velocity for two rotations $\omega =2\left(2\pi \right)\mathrm{rad}=4\pi \mathrm{rad}/\mathrm{s}$
spoke speed at the rim $=\omega r$
Speed of spokes at the axle $=0$
Average velocity $\left(v\right)=\frac{\omega r+0}{2}=\frac{\omega r}{2}$
Step 3. Calculations:
When we plug all the values into the equation above, we obtain
$e=\left(0.4\times {10}^{-4}\right)\times \left(\frac{1}{2}\times 4\pi \times 0.5\right)\times 0.5=6.28\times {10}^{-5} \mathrm{V}$

Hence, the correct answer is option C.