Question

The work done in turning a magnet of magnetic moment 'M' by an angle of 90$^\circ$ from the meridian is 'n' times the corresponding work done to turn it through an angle of 60$^\circ$ where the value of 'n' is ______.

• A. 0.5
• B. 2
• C. 0.25
• D. 1

Hint:

Rotating a magnet from $0^\circ$ to $60^\circ$ takes exactly HALF the energy required to rotate it all the way to $90^\circ$. This is a very common checkpoint calculation to have memorized!

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We must calculate the mechanical work done to rotate a magnetic dipole inside a uniform magnetic field for two different target angles, and then find the ratio ($n$) between those two amounts of work.

Step 2: Detailed Explanation:
The work done ($W$) in rotating a magnetic dipole of moment $M$ in a uniform magnetic field $B$ from an initial angle $\theta_1$ to a final angle $\theta_2$ is given by:
$W = \int_{\theta_1}^{\theta_2} \tau d\theta = MB (\cos \theta_1 - \cos \theta_2)$
The "magnetic meridian" is the resting equilibrium position where the magnet naturally aligns completely parallel with the external magnetic field. Therefore, the starting angle is always $\theta_1 = 0^\circ$.
Since $\cos(0^\circ) = 1$, the formula simplifies to:
$W = MB(1 - \cos \theta)$
Case 1: Turning through 90$^\circ$
Here, $\theta = 90^\circ$.
$W_{90} = MB(1 - \cos 90^\circ)$
Since $\cos 90^\circ = 0$:
$W_{90} = MB(1 - 0) = MB$ --- (Equation 1)
Case 2: Turning through 60$^\circ$
Here, $\theta = 60^\circ$.
$W_{60} = MB(1 - \cos 60^\circ)$
Since $\cos 60^\circ = 0.5$:
$W_{60} = MB(1 - 0.5) = 0.5 MB$ --- (Equation 2)
The problem states that $W_{90} = n \times W_{60}$.
Substitute the calculated values into this relationship:
$MB = n \times (0.5 MB)$
Cancel $MB$ from both sides:
$1 = 0.5 n$
$n = \frac{1}{0.5} = 2$

Step 3: Final Answer:
The value of 'n' is 2, matching option (b).