3. A magnetic dipole of dipole moment m is aligned parallel to an external magnetic field B. Work of 0.25 J has to be done in order to turn it through angle of 60°. Find the external counter torque that is required in or

Step-by-step solution

1. 1

   Given Data

   Initial angle $\theta_1 = 0^\circ$, final angle $\theta_2 = 60^\circ$, work done $W = 0.25\text{ J}$.
2. 2

   Work Done Formula

   Work done in rotating a magnetic dipole in a magnetic field is $W = mB(\cos\theta_1 - \cos\theta_2)$.

   $$W = mB(\cos\theta_1 - \cos\theta_2)$$
3. 3

   Calculate $mB$

   First, use the given work done to find the product $mB$.
4. 4

   Substitute Values

   Substitute given values into the work done formula: $0.25 = mB(\cos 0^\circ - \cos 60^\circ)$.

   $$0.25 = mB(\cos 0^\circ - \cos 60^\circ)$$
5. 5

   Simplify for $mB$

   Calculate $\cos 0^\circ = 1$ and $\cos 60^\circ = 1/2$. So $0.25 = mB(1 - 1/2) = mB/2$.

   $$0.25 = mB/2$$
6. 6

   Value of $mB$

   From $0.25 = mB/2$, we get $mB = 0.25 \times 2 = 0.5\text{ J}$.

   $$mB = 0.5\text{ J}$$
7. 7

   Torque Formula

   The torque on a magnetic dipole in a magnetic field is $\tau = mB\sin\theta$.

   $$\tau = mB\sin\theta$$
8. 8

   Calculate Torque

   At $\theta = 60^\circ$, $\tau = mB\sin 60^\circ = 0.5 \times \sqrt{3}/2$.

   $$\tau = 0.5 \times \frac{\sqrt{3}}{2}$$
9. 9

   Final Torque

   The external counter torque required is $\tau = 0.25\sqrt{3}\text{ J}$.

   $$\tau = 0.25\sqrt{3}\text{ J}$$