A conducting metal circular-wire-loop of radius $$r$$ is placed perpendicular to a magnetic field which varies with time as $$B = B_0 e^{-t/\tau}$$, where $$B_0$$ and $$\tau$$ are constants at time $$t = 0$$. If the resistance of the loop is $$R$$, then the heat generated in the loop after a long time $$(t \to \infty)$$ is

First we note that a changing magnetic flux through the loop induces an emf. The quantitative relation is given by Faraday’s law of electromagnetic induction, stated as

$$|\mathcal E| = \left|\frac{d\Phi}{dt}\right|,$$

where $$\mathcal E$$ is the induced emf and $$\Phi$$ is the magnetic flux through the loop.

The area of the given circular loop is

$$A = \pi r^{2}.$$

The magnetic field is time-dependent: $$B(t)=B_{0}e^{-t/\tau}.$$ Hence the flux through the loop at any instant $$t$$ is

$$\Phi(t)=B(t)\,A =B_{0}e^{-t/\tau}\,\pi r^{2}.$$

We now differentiate this flux with respect to time in order to find the emf:

$$\frac{d\Phi}{dt} =\pi r^{2}B_{0}\,\frac{d}{dt}\!\left(e^{-t/\tau}\right).$$

Using the elementary derivative

$$\frac{d}{dt}\!\left(e^{-t/\tau}\right)=-\frac{1}{\tau}e^{-t/\tau},$$

we get

$$\frac{d\Phi}{dt} =-\pi r^{2}B_{0}\,\frac{1}{\tau}e^{-t/\tau}.$$

Taking the magnitude (sign is irrelevant for power and heat), the induced emf becomes

$$\mathcal E(t)=\left|\frac{d\Phi}{dt}\right| =\frac{\pi r^{2}B_{0}}{\tau}\,e^{-t/\tau}.$$

The loop has resistance $$R$$, so by Ohm’s law the induced current is

$$I(t)=\frac{\mathcal E(t)}{R} =\frac{\pi r^{2}B_{0}}{\tau R}\,e^{-t/\tau}.$$

The instantaneous power dissipated as heat in the resistor is given by

$$P(t)=I^{2}(t)\,R.$$

Substituting the expression for $$I(t)$$, we obtain

$$P(t)=\left(\frac{\pi r^{2}B_{0}}{\tau R}\,e^{-t/\tau}\right)^{2}R =\frac{\pi^{2}r^{4}B_{0}^{2}}{\tau^{2}R}\,e^{-2t/\tau}.$$

To find the total heat $$Q$$ generated from $$t=0$$ to $$t\rightarrow\infty$$, we integrate this power over time:

$$Q=\int_{0}^{\infty}P(t)\,dt =\frac{\pi^{2}r^{4}B_{0}^{2}}{\tau^{2}R} \int_{0}^{\infty}e^{-2t/\tau}\,dt.$$

The standard integral

$$\int_{0}^{\infty}e^{-kt}\,dt=\frac{1}{k}$$

gives, with $$k=\frac{2}{\tau},$$

$$\int_{0}^{\infty}e^{-2t/\tau}\,dt =\frac{1}{2/\tau} =\frac{\tau}{2}.$$

Substituting this back, we have

$$Q=\frac{\pi^{2}r^{4}B_{0}^{2}}{\tau^{2}R}\,\frac{\tau}{2} =\frac{\pi^{2}r^{4}B_{0}^{2}}{2\tau R}.$$

Hence, the correct answer is Option B.