Question

A wire in the form of semi-circle of radius \( r \) rotates about its diameter with angular velocity \( \omega \) in a magnetic field \( B \). The axis of rotation is perpendicular to the field. The total resistance of the circuit is \( R \). If the mean power generated per period of rotation is \( \frac{(B\pi r^2 \omega)^2}{xR} \), then the value of \( x \) is

Hint:

Average of \( \sin^2 \omega t \) over a cycle is \( \frac{1}{2} \). Always use it in AC power problems.

Answer 1

Concept: Induced emf in rotating loop: \[ \varepsilon = B A \omega \sin \omega t \] Average power: \[ P_{\text{avg}} = \frac{\varepsilon_0^2}{2R} \]

Step 1: Maximum emf.
Area of semicircle: \[ A = \frac{1}{2} \pi r^2 \] \[ \varepsilon_0 = B \cdot \frac{1}{2} \pi r^2 \cdot \omega \]

Step 2: Mean power.
\[ P_{\text{avg}} = \frac{\varepsilon_0^2}{2R} = \frac{\left(B \cdot \frac{1}{2} \pi r^2 \omega\right)^2}{2R} \]

Step 3: Simplification.
\[ P_{\text{avg}} = \frac{(B \pi r^2 \omega)^2}{8R} \] Comparing with: \[ \frac{(B\pi r^2 \omega)^2}{xR} \Rightarrow x = 8 \]