QNEET-2024-Ph-57: NEET 2024 Physics - Class 12 NEET-2024-Ph-57 Solution: An iron bar of length L has magnetic moment M. It is bent at…

The magnetic moment of a straight bar magnet of length $L$ and pole strength $m$ is given by:

When the bar of total length $L$ is bent at its middle point:

1. The pole strength $m$ remains unchanged.
2. 
   
3. Each half-arm has a length of $l_1 = l_2 = \frac{L}{2}$.
4. 
   
5. The angle between the two arms is $\theta = 60^\circ$.
6. 
   

Step-by-Step Calculation

1. Finding the new effective length ($L'$):

The effective magnetic length is the straight-line displacement between the two endpoints (poles).

Using the law of cosines for a triangle with two sides of length $\frac{L}{2}$ and an included angle of $60^\circ$:

Since $\cos 60^\circ = \frac{1}{2}$:

(Alternatively, because both arms are equal in length and the angle between them is $60^\circ$, the triangle formed by the two arms and the line connecting the two poles is an equilateral triangle. Therefore, the distance between the two ends is simply $\frac{L}{2}$.)

2. Calculating the new magnetic moment ($M'$):

Substituting $m \cdot L = M$:

Answer:

The magnetic moment of the new magnet is $\frac{M}{2}$ (or $0.5\,M$).