Animated Solution for Physics - Magnetic Effects of Current: A magnet of total magnetic moment $10^{-2}\hat{i}{\text{ A-m}}^2$ is placed in a time varying magnetic field, $B\hat{i}(\cos \omega t)$, where $B=1\text{T}$ and $\omega =0.125\text{ rad/s}$. The work done for reversing the direction of the magnetic moment at $t=1\text{s}$ is

Have you ever solved a physics problem perfectly, only to find that your answer doesn't match any of the options? It is a heart-sinking moment. But what if I told you that sometimes, the error isn't in your physics, but in the question itself?

Today, we are going to dive into a fascinating JEE Main problem that tests not only your grasp of electromagnetism but also your analytical intuition to spot a hidden typo. Let's unravel the mystery of the reversing magnetic dipole!

A Dipole in a Fluctuating Field

Imagine a tiny bar magnet, a magnetic dipole, resting in a region of space. This isn't just any static magnetic field; it's a time-varying field that pulses rhythmically, described by the equation $\mathbf{B}(t)=B\cos (\omega t)\hat{i}$.

Our dipole has a magnetic moment $\mathbf{M}=10^{-2}\hat{i}{\text{ A-m}}^2$. Notice the $\hat{i}$ vector? This tells us that initially, the dipole is perfectly aligned with the magnetic field along the x-axis. It is in a state of stable equilibrium, resting comfortably at the bottom of its potential energy well.

But the problem asks us to disturb this peace. We need to find the work done to reverse the direction of the magnetic moment at the exact instant $t=1\text{ s}$.

Work and Potential Energy

To twist a magnet against the invisible grip of a magnetic field, an external agent must do work. This work doesn't just vanish; it is stored in the system as magnetic potential energy.

The potential energy $U$ of a magnetic dipole in a uniform magnetic field is given by the elegant dot product:

Where $\theta$ is the angle between the magnetic moment and the magnetic field.

The work done by an external agent to rotate the dipole from an initial angle ${\theta }_1$ to a final angle ${\theta }_2$ is simply the change in this potential energy:

In our scenario, "reversing" the dipole means rotating it from its perfectly aligned state (${\theta }_1=0^{\circ }$) to a completely anti-parallel state (${\theta }_2=180^{\circ }$). Let's plug these angles into our master equation:

This makes perfect physical sense. We are taking the dipole from its lowest energy state ($-MB$) to its highest energy state ($+MB$). The total energy climb is exactly $2MB$.

A Typo in the Matrix

Now, we need to evaluate the magnetic field $B$ at the specific instant $t=1\text{ s}$. The problem states that $B_0=1\text{ T}$ and the angular frequency $\omega =0.125\text{ rad/s}$.

Let's blindly follow the numbers:

If you punch $\cos (0.125\text{ rad})$ into a calculator, you get approximately $0.992\text{ T}$.

Substituting this back into our work equation:

Look at the options: (a) 0.01 J, (b) 0.007 J, (c) 0.014 J, (d) 0.028 J.

Our calculated value of $0.01984\text{ J}$ is nowhere to be found! The official solution provided in many textbooks bizarrely claims that $0.0198\text{ J}$ is "nearest to 0.014 J", which is mathematically absurd (it's clearly closer to 0.02 J).

Are we wrong? Is physics broken? No. It's time to think like an examiner.

Decoding the True Intent

In competitive exams, numbers are rarely random. The value $0.125$ is a very specific fraction: it is exactly $\frac{1}{8}$.

What if the unit "rad/s" was a typographical error? What if the examiner intended for the frequency to be $\frac{1}{8}$ of a revolution per second (rev/s) or Hertz?

Let's test this hypothesis. If the frequency $f=0.125\text{ rev/s}$, then the angular frequency $\omega$ in radians per second would be:

This is a beautiful, standard angle! Let's see what happens if we use $\omega =\frac{\pi }{4}\text{ rad/s}$.

Bringing It All Together

Let's recalculate the magnetic field at $t=1\text{ s}$ using our newly decoded angular frequency:

We know that $\cos (45^{\circ })$ or $\cos (\frac{\pi }{4})$ is exactly $\frac{1}{\sqrt{2}}$.

Now, let's substitute this pristine value back into our work equation:

Since $\sqrt{2}\approx 1.414$, we get:

Rounding to three decimal places, we get exactly 0.014 J.

Look at option (c). It is a perfect match!

By trusting our physics and applying a bit of analytical deduction, we didn't just solve the problem; we debugged it. This is the hallmark of a true physicist—not just crunching numbers, but understanding the harmony and intent behind them. Keep this mindset, and no tricky question will ever stand in your way!