1. Given data

• Radius of coil: $r = 1\,\text{m}$
• Resistance: $R = 2\,\mu\Omega = 2\times 10^{-6}\,\Omega$
• Magnetic field: $B(t)=\frac{4}{\pi}\times 10^{-3}\left(1-\frac{t}{100}\right)\,\text{T}$

The field is perpendicular to the plane of the coil.

It becomes zero when $1-\frac{t}{100}=0 \Rightarrow t=100\,\text{s}$

So we need the energy dissipated from $t=0$ to $t=100\,\text{s}$.

2. Magnetic flux through the coil

Area of the circular coil: $A=\pi r^2=\pi(1)^2=\pi\,\text{m}^2$

Flux: $\Phi = BA = B\pi$

Substitute $B$: $\Phi(t)=\pi\cdot \frac{4}{\pi}\times 10^{-3}\left(1-\frac{t}{100}\right)$ $\Phi(t)=4\times 10^{-3}\left(1-\frac{t}{100}\right)\,\text{Wb}$

3. Induced emf

By Faraday's law, $|\mathcal E|=\left|\frac{d\Phi}{dt}\right|$

Differentiate: $\Phi(t)=4\times 10^{-3}\left(1-\frac{t}{100}\right)$ $\frac{d\Phi}{dt}=4\times 10^{-3}\left(-\frac{1}{100}\right)=-4\times 10^{-5}\,\text{V}$

Hence, $\mathcal E = 4\times 10^{-5}\,\text{V}$

This is constant.

4. Current in the coil

Using Ohm’s law, $I=\frac{\mathcal E}{R}=\frac{4\times 10^{-5}}{2\times 10^{-6}}=20\,\text{A}$

5. Power dissipated

$P=I^2R=(20)^2(2\times 10^{-6})$ $P=400\times 2\times 10^{-6}=8\times 10^{-4}\,\text{W}$

Since power is constant, total energy dissipated in $100\,\text{s}$ is $E=Pt=(8\times 10^{-4})(100)=8\times 10^{-2}\,\text{J}$

$E=0.08\,\text{J}=80\,\text{mJ}$

6. Final answer

$\boxed{80}$

The derived answer matches the stored correct answer.