If 50.0 cmof copper wire (diameter = 1.00 mm) 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?

i) Length of wire, L = 50 cm

ii) Diameter of wire, d = 1mm

iii) Rate of increase of magnetic field,$\frac{fB}{dt}=10\mathrm{mT}/\mathrm{s}$

First, find the emf induced in the wire. By substituting the emf induced and the resistance of the wire in the formula for rate of energy dissipated, find the rate of thermal energy generated in the loop.

Faraday's law of electromagnetic inductionstates, Whenever a conductor is placed in a varying magnetic field, an electromotive force is induced in it.

Formulae are as follow:

$$\begin{gathered} \left|\mathrm{\epsilon }\right|=\frac{{\mathrm{d\varphi }}_{\mathrm{B}}}{dt} \\ \mathrm{\varphi }=BA \\ \mathrm{P}=\frac{{\mathrm{\epsilon }}^2}{\mathrm{R}} \\ R=\frac{\rho L}{A} \end{gathered}$$

Where,$\mathrm{\Phi }$is magnetic flux, B is magnetic field, A is area, R is resistance,𝜀 is emf, L is length, P is power,𝜌 is resistivity.

The rate of thermal energy generated in the loop is given by the equations 26-28,

$P=\frac{{\epsilon }^2}{R}.......................................\left(1\right)$

Where, R is the resistance of the wire and is given by,

$R=\frac{\rho L}{A}$

The area enclosed by the loop is,

$A_{loop}=\tau \tau r_{loop}^2=\tau \tau \frac{d^2}{4}$.

So the resistance becomes,

role="math" localid="1661761767125" $R=\frac{\rho L}{\tau \tau {\displaystyle \frac{d^2}{4}}}....................................................\left(2\right)$

The length of the wire is,

$L=2\tau \tau r$

Rearranging this equation for radius,

$r=\frac{L}{2\tau \tau }$

So area can also be written as,

$A_{loop}=\tau \tau r_{loop}^2=\tau \tau {\left(\frac{L}{2\tau \tau }\right)}^2$

Now, the magnitude of induced emf in the loop is given by Faraday’s law,

$\left|\mathrm{\epsilon }\right|=\frac{d{\mathrm{\varphi }}_{\mathrm{B}}}{dt}$

Since,${\mathrm{\varphi }}_{\mathrm{B}}=B{A}_{loop}=B\tau \tau {\left(\frac{L}{2\tau \tau }\right)}^2$

Therefore,

$\left|\mathrm{\epsilon }\right|=\frac{L^2}{4\tau \tau }\frac{dB}{dt}............................................\left(3\right)$

Substituting the expression for resistance and induced emf infrom (2) and (3),

$$\begin{gathered} P=\frac{{\epsilon }^2}{R}=\frac{{\left({\displaystyle \frac{L^2}{4\tau \tau }\frac{dB}{dt}}\right)}^2}{\rho L/\left({\displaystyle \frac{\tau \tau d^2}{4}}\right)} \\ P=\frac{d^2{L}^3}{64\tau \tau \rho }{\left(\frac{dB}{dt}\right)}^2 \end{gathered}$$

But the rate of change of magnetic field is given as$\frac{dB}{dt}=0.0100\mathrm{T}/\mathrm{s}$

Also,$\rho =1.69\times {10}^{-8}\Omega .m$is the resistivity of copper wire, from tables 27-1

The rate of thermal energy generated in the loop is,

$$\begin{gathered} P=\frac{{\left(1\times {10}^{-3}\mathrm{m}\right)}^2{\left(0.500\mathrm{m}\right)}^3}{64\mathrm{\tau \tau }\left(1.69\times {10}^{-8}\mathrm{\Omega }.\mathrm{m}\right)}{\left(0.0100\frac{\mathrm{T}}{\mathrm{s}}\right)}^2 \\ P=3.68\times {10}^{-6}\mathrm{W} \end{gathered}$$

Hence, the rate of thermal energy generated in the loop is$3.68\times {10}^{-6}\mathrm{W}$

Therefore, by using the concept of induced emf and it’s formula, the rate of thermal energy can be determined.