QJEEM19Phe5d3: JEE Main 2019 Physics - Class 12 JEEM19Phe5d3 Solution: A magnet of total magnetic moment 10 -2 i ^ A-m 2 is placed in…

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 $B_{\text{RMS}}$ :

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