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 ______.
Show 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!
Step 1: Understanding the Concept:
The work done in rotating a magnet in a magnetic field $B$ from angle $\theta_1$ to $\theta_2$ is $W = MB(\cos \theta_1 - \cos \theta_2)$. Starting from the meridian means $\theta_1 = 0^\circ$. Step 2: Formula Application:
$W = MB(1 - \cos \theta)$.
For 90°: $W_1 = MB(1 - \cos 90^\circ) = MB(1 - 0) = MB$.
For 60°: $W_2 = MB(1 - \cos 60^\circ) = MB(1 - 0.5) = 0.5 MB$. Step 3: Explanation:
Given $W_1 = n W_2$.
$MB = n (0.5 MB) \implies n = \frac{1}{0.5} = 2$. Step 4: Final Answer:
The value of $n$ is 2.