A thin rod of length L is magnetised and has magnetic moment M.The rod is then bent in a semicircular arc.The magnetic moment in the new shape is

Options

A. `(M)/(L)`
B. `(M)/(pi)`
C. `(M)/(2pi)`
D. `(2M)/(pi)`

Explanation

To solve the problem of finding the magnetic moment of a thin rod bent into a semicircular arc, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Initial Magnetic Moment**:
The magnetic moment \( M \) of the thin rod is given by the formula:
\[ M = P \times L \]
where \( P \) is the pole strength and \( L \) is the length of the rod.

2. **Determine the Shape After Bending**:
When the rod is bent into a semicircular arc, the length of the rod remains the same, which is \( L \). The circumference of a semicircle is given by:
\[ \text{Circumference} = \pi R \]
where \( R \) is the radius of the semicircle.

3. **Relate Length to Radius**:
Since the length of the rod is equal to the semicircular arc, we have:
\[ \pi R = L \]
From this, we can solve for the radius \( R \):
\[ R = \frac{L}{\pi} \]

4. **Find the Distance Between the Poles**:
In a semicircular shape, the distance between the north and south poles (the effective length for magnetic moment calculation) is twice the radius:
\[ \text{Distance between poles} = 2R = 2 \times \frac{L}{\pi} = \frac{2L}{\pi} \]

5. **Calculate the New Magnetic Moment**:
The new magnetic moment \( M' \) for the semicircular arc can be calculated using the pole strength \( P \) and the new effective length:
\[ M' = P \times \left(\frac{2L}{\pi}\right) \]
Since \( M = P \times L \), we can express \( P \) in terms of \( M \):
\[ P = \frac{M}{L} \]
Substituting this back into the equation for \( M' \):
\[ M' = \left(\frac{M}{L}\right) \times \left(\frac{2L}{\pi}\right) = \frac{2M}{\pi} \]

### Final Answer:
The magnetic moment in the new shape (semicircular arc) is:
\[ M' = \frac{2M}{\pi} \]