From JEE Advanced 2022: Rotating Solenoid

One of the finest questions of the 2022 JEE Advanced was unfortunately dropped in the evaluation due to a small error. But since the error is a very superficial one, the question remains open for us to enjoy and learn from in its original form.

It was Q.15 in Paper-1, the first of the list match questions. Read it and see if you can spot the error.

The question is special because it forces us to write all the relevant quantities in vector form. This makes it one of a handful of questions that the JEE paper setters include every year that gradually but surely extend the scope of the exam, by bringing into focus skills that were almost never needed to solve the papers in previous years.

Of course, we can do this problem avoiding vector notation completely (and instead proceeding carefully using only the right hand rule for directions), but it would be painful. I know, I tried it and solved it wrong. Twice. So we will do it in the proper way, writing everything vectorially.

Make sure you have tried your best at solving before reading further.

As the solenoid changes orientation, its magnetic field (which is always axial) changes direction too. This variation is given in terms of the axially pointing unit vector n. It is here that a small, rather insignificant error crept in: none of the vectors in the first column are unit vectors! They all have an extra factor of 1/sqrt(2). Don't worry if you could not find it; a great many people I know who should have caught it did not either.

So, as the axial magnetic field changes direction in the given way, a current is induced in the circular loop, which is fixed. This induced current produces its own field, and hence gives the loop its magnetic moment μ. To reach the goal of finding the instantaneous torque on the loop due to the solenoid's field, we must first write the magnetic moment as a vector, and then find the torque using τ = μ x B.

Now, because the same problem has to be solved 4 times here, we must be smart from the beginning and write everything in an efficient way. The steps to be followed are:

1. write the magnetic field of the solenoid using the well-known formula
   
   
   
   
   
2. the magnetic moment μ of the loop is found simply from dB_z/dt using Faraday's law, where B_z is the component of the field normal to the loop's plane (only this component creates flux through the loop). The negative sign is due to Lenz's law
   
   
   
   Do not be fazed by the B.z - that is just how we write the z-component of B. We could have simply selected out the z-component by looking at B and taken its derivative. In fact, that is what we will do in practice. And the second z at the end is to make the magnetic moment a vector.
3. then find the torque using τ = m x B. Here, B is the net field and not just the component normal to the loop's plane. Must be careful!
4. plug in the given time instant

As any good problem should be, this one is instructive. It uses all the important concepts of electromagnetic induction, but above and beyond that, it forces us to write general expressions in their vector form, and thus to make efficient use of the concepts learned. If the problem is worked out multiple times over a span of time, it crystallizes these new skills. Of course, the basics of induction must be known and well practiced beforehand.

If the terms in list I are made unit vectors by deleting the leading factor, and the given expression for α is changed to fit the correct results (the error in α was due to the error in n), the answers are as below: