Question

The work done in rotating a bar magnet which is initially in the direction of a uniform magnetic field through $ 45^\circ $ is $ W $. The additional work to be done to rotate the magnet further through $ 15^\circ $ is.

• A. $ \frac{W}{\sqrt{2}} $
• B. $ \frac{W}{2} $
• C. $ W\sqrt{2} $
• D. $ 2W $

Hint:

The work done in rotating a bar magnet depends on the cosine of the angle between the magnetic moment and the magnetic field. Use the formula $ W = M B (1 - \cos\theta) $ to relate the work to the angle.

Answer 1

The Correct Option is A

Step 1: Known Information. 
Initial orientation of the bar magnet: aligned with the magnetic field ($ \theta_0 = 0^\circ $).
Work done to rotate the magnet from $ 0^\circ $ to $ 45^\circ $ is $ W $.
We need to find the additional work required to rotate the magnet from $ 45^\circ $ to $ 60^\circ $ (an additional $ 15^\circ $). 
Step 2: Formula for Work Done in Rotating a Bar Magnet. 
The work done ($ W $) in rotating a bar magnet in a uniform magnetic field is given by: $$ W = M B (1 - \cos\theta) $$ where:
$ M $ is the magnetic moment of the magnet,
$ B $ is the magnetic field strength,
$ \theta $ is the angle between the magnetic moment and the magnetic field.
Step 3: Work Done to Rotate from $ 0^\circ $ to $ 45^\circ $. When the magnet is rotated from $ 0^\circ $ to $ 45^\circ $: $$ W_{0 \to 45} = M B (1 - \cos 45^\circ) $$ We know that: $$ \cos 45^\circ = \frac{1}{\sqrt{2}} $$ Thus: $$ W_{0 \to 45} = M B \left(1 - \frac{1}{\sqrt{2}}\right) = W $$ Step 4: Work Done to Rotate from $ 45^\circ $ to $ 60^\circ $.
When the magnet is rotated from $ 45^\circ $ to $ 60^\circ $, the total work done from $ 0^\circ $ to $ 60^\circ $ is: $$ W_{0 \to 60} = M B (1 - \cos 60^\circ) $$ We know that: $$ \cos 60^\circ = \frac{1}{2} $$ Thus: $$ W_{0 \to 60} = M B \left(1 - \frac{1}{2}\right) = M B \cdot \frac{1}{2} $$ The additional work done from $ 45^\circ $ to $ 60^\circ $ is: $$ W_{45 \to 60} = W_{0 \to 60} - W_{0 \to 45} $$ Substitute the expressions for $ W_{0 \to 60} $ and $ W_{0 \to 45} $: $$ W_{45 \to 60} = M B \cdot \frac{1}{2} - M B \left(1 - \frac{1}{\sqrt{2}}\right) $$ Simplify: $$ W_{45 \to 60} = M B \left(\frac{1}{2} - 1 + \frac{1}{\sqrt{2}}\right) $$ $$ W_{45 \to 60} = M B \left(-\frac{1}{2} + \frac{1}{\sqrt{2}}\right) $$ Step 5: Relate to Given Work $ W $.
From Step 3, we know: $$ W = M B \left(1 - \frac{1}{\sqrt{2}}\right) $$ Thus: $$ M B = \frac{W}{1 - \frac{1}{\sqrt{2}}} $$ Substitute $ M B $ into the expression for $ W_{45 \to 60} $: $$ W_{45 \to 60} = \frac{W}{1 - \frac{1}{\sqrt{2}}} \cdot \left(-\frac{1}{2} + \frac{1}{\sqrt{2}}\right) $$ Simplify the term $ -\frac{1}{2} + \frac{1}{\sqrt{2}} $: $$ -\frac{1}{2} + \frac{1}{\sqrt{2}} = \frac{-\sqrt{2} + 2}{2\sqrt{2}} = \frac{2 - \sqrt{2}}{2\sqrt{2}} $$ Thus: $$ W_{45 \to 60} = \frac{W}{1 - \frac{1}{\sqrt{2}}} \cdot \frac{2 - \sqrt{2}}{2\sqrt{2}} $$ Simplify $ 1 - \frac{1}{\sqrt{2}} $: $$ 1 - \frac{1}{\sqrt{2}} = \frac{\sqrt{2} - 1}{\sqrt{2}} $$ So: $$ W_{45 \to 60} = \frac{W}{\frac{\sqrt{2} - 1}{\sqrt{2}}} \cdot \frac{2 - \sqrt{2}}{2\sqrt{2}} $$ $$ W_{45 \to 60} = W \cdot \frac{\sqrt{2}}{\sqrt{2} - 1} \cdot \frac{2 - \sqrt{2}}{2\sqrt{2}} $$ Simplify: $$ W_{45 \to 60} = W \cdot \frac{2 - \sqrt{2}}{2(\sqrt{2} - 1)} $$ Rationalize the denominator: $$ \frac{2 - \sqrt{2}}{\sqrt{2} - 1} = \frac{(2 - \sqrt{2})(\sqrt{2} + 1)}{(\sqrt{2} - 1)(\sqrt{2} + 1)} = \frac{(2 - \sqrt{2})(\sqrt{2} + 1)}{2 - 1} = (2 - \sqrt{2})(\sqrt{2} + 1) $$ Expand: $$ (2 - \sqrt{2})(\sqrt{2} + 1) = 2\sqrt{2} + 2 - 2 - \sqrt{2} = \sqrt{2} $$ Thus: $$ W_{45 \to 60} = W \cdot \frac{\sqrt{2}}{2} $$ $$ W_{45 \to 60} = \frac{W}{\sqrt{2}} $$ Final Answer: $ \boxed{\frac{W}{\sqrt{2}}} $