Question

A rectangular wire loop of sides $8\, cm$ and $2\, cm$ with a small cut is moving out of a region of uniform magnetic field of magnitude $0.3 \, T$ directed normal to the loop. Suppose the loop is stationary but the current feeding the electromagnet that produces the magnetic field is gradually reduced so that the field decreases from its initial value of $0.3 \, T$ at the rate of $0.02 \, T / s$. If the cut is joined and the loop has a resistance of $1.6 \, \Omega$, how much power is dissipated by the loop as heat?

• A. $6.4 \times 10^{10} \, W$
• B. $0.64 \times 10^{10} \, $
• C. $64 \times 10^{10} \, W$
• D. $6.4 \times 10^{-10} \, W$

Answer 1

Here $d B / d t=0.02 \,T / s$,
Area of the loop,
$A=l \times b=8 \, cm \times 2 \, cm $
$=16 \,cm ^{2}=16 \times 10^{-4} \,m ^{2}$
Resistance of the loop, $R=1.6 \,\Omega$
Let $\varepsilon$ be the magnitude of the induced emf in the loop.
Clearly, $\varepsilon=\frac{d \phi_{B}}{d t}=\frac{d}{d t}\left(\phi_{B}\right)$
$ \Rightarrow \frac{d}{d t}(B A)=A\left(\frac{d B}{d t}\right)$
(as $\phi_{B}=B A$ and $A$ is a constant)
or $\varepsilon=\left(16 \times 10^{-4} m ^{2}\right)(0.02 T / s )=32 \times 10^{-6} \,V$
When the cut in the loop is joined, power dissipated
$P=\frac{\varepsilon^{2}}{R}=\frac{\left(32 \times 10^{-6} V \right)^{2}}{1.6 \, \Omega}$
$=6.4 \times 10^{-10} \,W$