Question

Consider two loops of currents. The first is fixed on an horizontal plane and its axis points up (i.e. it rotates counterclockwise if seen from above). The second loop has its center fixed on the same plane, but is free to rotate. What will be the final equilibrium position of the second loop once let go?

(For an Electromagnetism class using the Introduction to Electrodynamics, David Griffiths 3rd edition textbook)

Answer 1

Two different principles are at play here: we need to know the direc- tion of the magnetic field created by one of the wires, and then what direction the force on the other wire due to that field is. For the first part, we need to use the Biot-Savart law:

$\bold{B} =\frac{\mu _{0}}{4\pi}\bold{I} \frac{d\bold{I} \times \bold{\hat{r}}}{r^{2}}$

Let’s think about the first loop. Using screw rule, the magnetic field is directed outward of the page. This magnetic field will effect the second loop and the second loop will try to oppose the induced magnetic field. For that the second loop will produce current in clockwise direction, so that magnetic field directed into the page will be created. And so the force of interaction between the two loops will be repulsive in nature.

Since first loop is in anti-clockwise direction, the rotation of second loop will be in clockwise direction. Using Lorentz law, you can prove this. ie

$\bold{F} = \bold{I}dl \times \bold{B}$

The second loop is in the influence of a magnetic field directed outward of the page. But the Magnetic moment produced by the second loop will be directed into the page(using thumb rule).

So the torque of the second loop is given as

$\tau = \bold{M}\times\bold{B} = \bold{M}\bold{B}sin\Theta = 0$

The interaction energy of the second loop is given by,

$U = -\bold{M}\cdot \bold{B} = \bold{M} \bold{B}$

This is the maximum energy that the second loop can have and this loop will be in unstable equilibrium. And loop will finally come to equilibrium, when the magnetic moment produced by the loop and the magnetic field are in parallel to each other, so that,

U = -MBcos0 = -MB

This is the minimum energy the loop can have and will be in stable equilibrium state.

Inorder to achieve it, the current in second loop must flow similar to 1st loop. Then only the second loop can achieve equilibrium.