Question

A copper wire of length 50.0 cm and total resistance of $1.1 \times 10^{-2} \Omega$ is formed into a circular loop and placed perpendicular to a uniform magnetic field that is increasing at the constant rate of $10.0\, mT / s$. At what rate is thermal energy generated in the loop?

• A. $1.32 \times 10^{-8} W$
• B. $2.36 \times 10^{-4} W$
• C. $3.62 \times 10^{-6} W$
• D. $4.23 \times 10^{-5} W$

Answer 1

Answer: (C)

Here, $L=50.0\, cm =50 \times 10^{-2}\, m$
$R=1.1 \times 10^{-2} \Omega$
$\frac{d B}{d t}=10.0\, mT / s =10 \times 10^{-3} T / s$
Let $r$ be the radius of a circular loop.
$\therefore 2 \pi r=L,\, r=\frac{L}{2 \pi}=\frac{50 \times 10^{-2}}{2 \pi} m =\frac{1}{4 \pi} m$
Magnetic flux linked with the loop is
$\phi=B A \cos \theta=B \pi r^{2} \cos 0^{\circ}=B \pi r^{2} \left(\because \theta=0^{\circ}\right)$
Emf induced in the loop is $|\varepsilon|=\frac{d \phi}{d t}=\pi r^{2} \frac{d B}{d t}$
Rate at which thermal energy generated in the loop is
$P=\frac{|\varepsilon|^{2}}{R}=\frac{\left(\pi r^{2}\right)^{2}\left(\frac{d B}{d t}\right)^{2}}{R}$
Substituting the given values, we get
$P=\frac{\pi^{2} \times\left(\frac{1}{4 \pi}\right)^{4} \times\left(10 \times 10^{-3}\right)^{2}}{1.1 \times 10^{-2}}=3.62 \times 10^{-6} W$