A square loop of side $a=0.1\text{ m}$ and resistance $R=0.5\Omega$ is moved out of a magnetic field $B=1\text{ T}$ at a constant velocity $v=2\text{ m/s}$. What is the power $P$ dissipated as heat in the loop?

Solution

1. Identify the given values: side length $a=0.1\text{ m}$, magnetic field $B=1\text{ T}$, velocity $v=2\text{ m/s}$, and resistance $R=0.5\Omega$.
2. Calculate the induced EMF using the formula $$\begin{array}{rl}\epsilon & =Bav\\ & =1\times 0.1\times 2\\ & =0.2\text{ V}\end{array}$$.
3. Determine the power dissipated as heat using $$\begin{array}{rl}P& =\frac{{\epsilon }^2}{R}\\ & =\frac{(0.2{)}^2}{0.5}\end{array}$$.
4. Simplify the expression to find $$\begin{array}{rl}P& =\frac{0.04}{0.5}\\ & =0.08\text{ W}\end{array}$$.
5. Hence the answer is (A).