Question

When a bar magnet placed parallel to the magnetic field is rotated by $45^\circ$, the amount of work done is $2.07J$. The amount of work to be done to rotate the magnet further by $45^\circ$ is

• A. $2.07J$
• B. $3J$
• C. $4.41J$
• D. $5J$
• E. $6.21J$

Hint:

Work depends on cosine difference of angles in magnetic rotation.

Answer 1

The Correct Option is D

Concept:
• Work done: \[ W = MB(\cos\theta_1 - \cos\theta_2) \]

Step 1: First rotation $0^\circ \to 45^\circ$
\[ W_1 = MB(1 - \cos45^\circ) = MB(1 - \tfrac{1}{\sqrt{2}}) \] Given: \[ W_1 = 2.07J \]

Step 2: Second rotation $45^\circ \to 90^\circ$
\[ W_2 = MB(\cos45^\circ - 0) = MB\left(\tfrac{1}{\sqrt{2}}\right) \]

Step 3: Ratio
\[ \frac{W_2}{W_1} = \frac{1/\sqrt{2}}{1 - 1/\sqrt{2}} = \frac{0.707}{0.293} \approx 2.414 \] \[ W_2 = 2.07 \times 2.414 \approx 5J \] Final Conclusion:
Option (D)