1. Given data

• Magnetic moment of the magnet: $\vec m = 10^{-2}\,\hat i\; \text{A m}^2$ so $m = 10^{-2}\;\text{A m}^2$

• Time-varying magnetic field: $\vec B(t) = B\cos(\omega t)\,\hat i$ with $B=1\;\text{T}, \qquad \omega = 0.125\;\text{rad s}^{-1}$

• We need the work done in reversing the magnetic moment at $t=1\,$s.

2. Potential energy of a magnetic dipole

The potential energy of a magnetic dipole in a magnetic field is $U = -\vec m\cdot \vec B = -mB\cos\theta$ where $\theta$ is the angle between $\vec m$ and $\vec B$.

Initially, $\vec m$ and $\vec B$ are along the same direction ($\hat i$), so $\theta_i = 0$ Hence, $U_i = -mB(t)$

After reversal, the magnetic moment becomes opposite to the field direction, so $\theta_f = \pi$ Thus, $U_f = +mB(t)$

Therefore, work done in reversing the dipole is the increase in potential energy: $W = U_f - U_i = mB(t)-(-mB(t)) = 2mB(t)$

3. Evaluate magnetic field at $t=1$ s

$B(t)=B\cos(\omega t)=1\cdot \cos(0.125\times 1)$ $B(1)=\cos(0.125)\;\text{T}$

Now, $\cos(0.125) \approx 0.9922$

So, $B(1) \approx 0.9922\;\text{T}$

4. Compute the work done

$W=2mB(1)$ $W=2\times 10^{-2}\times 0.9922$ $W\approx 0.01984\;\text{J}$

So the required work is approximately $\boxed{0.020\;\text{J}}$

5. Check with the options

Given options are:

• A: $0.014\,$J
• B: $0.028\,$J
• C: $0.01\,$J
• D: $0.007\,$J

Our calculated value $0.020\,$J does not match any option.

6. Comparison with stored correct answer

Stored correct answer is A: $0.014\,$J.

But from standard formula, $W = 2mB\cos(\omega t)$ at $t=1\,$s gives $W \approx 0.01984\,\text{J}$ not $0.014\,$J.

So the stored answer appears inconsistent with the given data. It is possible there is a misprint in the question/options (for example, a different value of $\omega$, $t$, or magnetic moment may have been intended).