JEE Main 2019 — Magnetic Properties Of Matter Question with Solution

To determine the work done in reversing the direction of a magnetic moment in a time-varying magnetic field, we'll follow these steps:

Given:

Magnetic moment: $\mathbf{m}=10^{-2}\hat{i}$ A·m²

Magnetic field: $\mathbf{B}(t)=B\cos (\omega t)\hat{i}$, where $B=1$ T and $\omega =0.125$ rad/s

Time at which reversal occurs: $t=1$ s

Step 1: Calculate the Magnetic Field at $t=1$ s

$B(t)=B\cos (\omega t)=1\times \cos (0.125\times 1)=\cos (0.125\text{ rad})$

Compute $\cos (0.125\text{ rad})$:

$\cos (0.125\text{ rad})\approx 0.9922$

So,

$B(t)\approx 0.9922\text{ T}$

Step 2: Calculate the Work Done

The potential energy $U$ of a magnetic dipole in a magnetic field is:

$U=-\mathbf{m}\cdot \mathbf{B}=-mB\cos \theta$

The work done $W$ in reversing the magnetic moment from $\theta =0^{\circ }$ to $\theta =180^{\circ }$ is:

$W=U_{\text{final}}-U_{\text{initial}}=[-mB\cos (180^{\circ })]-[-mB\cos (0^{\circ })]=mB(\cos 0^{\circ }-\cos 180^{\circ })$

Simplify:

$W=mB(1-(-1))=2mB$

Substitute the values:

$W=2\times (10^{-2}{\text{ A⋅m}}^2)\times (0.9922\text{ T})\approx 0.01984\text{ J}$

Step 3: Approximate Using RMS Value

Given that the magnetic field is time-varying, we can consider the root mean square (RMS) value of $\cos (\omega t)$ over a complete cycle:

$\text{RMS of }\cos (\omega t)=\frac{1}{\sqrt{2}}$

Thus, the RMS value of the magnetic field is:

$B_{\text{RMS}}=B\times \frac{1}{\sqrt{2}}=1\times \frac{1}{\sqrt{2}}\approx 0.7071\text{ T}$

Now, calculate the work done using $B_{\text{RMS}}$:

$W_{\text{RMS}}=2m{B}_{\text{RMS}}=2\times (10^{-2})\times 0.7071\approx 0.01414\text{ J}$