A circular coil of radius $r=0.1\text{ m}$ and $N=500$ turns is placed in a uniform magnetic field $B=0.5\text{ T}$ such that the plane of the coil is perpendicular to the field. If the coil is rotated by $180^{\circ }$ about its diameter in $\Delta t=0.2\text{ s}$, what is the average induced electromotive force $\epsilon$ in the coil?

Solution

1. Identify the given values: $N=500$, $r=0.1\text{ m}$, $B=0.5\text{ T}$, and $\Delta t=0.2\text{ s}$.
2. Calculate the area of the coil: $$\begin{array}{rl}A& =\pi r^2\\ & =\pi (0.1{)}^2\\ & =0.01\pi {\text{ m}}^2\end{array}$$.
3. Determine the change in magnetic flux: $$\begin{array}{rl}\Delta \Phi & =BA\cos (180^{\circ })-BA\cos (0^{\circ })\\ & =-2BA\\ & =-2(0.5)(0.01\pi )\\ & =-0.01\pi \text{ Wb}\end{array}$$.
4. Apply Faraday's Law of Induction: $$\begin{array}{rl}\epsilon & =-N\frac{\Delta \Phi }{\Delta t}\\ & =-500\times \frac{-0.01\pi }{0.2}\\ & =2500\times 0.01\pi \\ & =25\pi \approx 1.57\text{ V}\end{array}$$.
5. Hence the answer is (A).