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 $10.0mT/s$. 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, $L=50.0cm=50\times 10^{-2}m$
$R=1.1\times 10^{-2}\Omega$
$\frac{dB}{dt}=10.0mT/s=10\times 10^{-3}T/s$
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
$\varphi =BA\cos \theta =B\pi r^2\cos 0^{\circ }=B\pi r^2\left(\because \theta =0^{\circ }\right)$
Emf induced in the loop is $|\epsilon |=\frac{d\varphi }{dt}=\pi r^2\frac{dB}{dt}$
Rate at which thermal energy generated in the loop is
$P=\frac{|\epsilon {|}^2}{R}=\frac{{\left(\pi r^2\right)}^2{\left(\frac{dB}{dt}\right)}^2}{R}$
Substituting the given values, we get
$P=\frac{{\pi }^2\times {\left(\frac{1}{4\pi }\right)}^4\times {\left(10\times 10^{-3}\right)}^2}{1.1\times 10^{-2}}=3.62\times 10^{-6}W$