The magnetic moment of a magnetised steel wire is M. If wire is bent to form a semin-circular arc then magnetic moment becomes

Options

A. `pi/2m`
B. `pi m/2`
C. `2m/pi`
D. `m/4 pi`

Answer

To solve the problem, we need to determine how the magnetic moment changes when a magnetized steel wire is bent into a semicircular arc. Let's break down the solution step-by-step.

### Step-by-Step Solution:

1. **Understanding Magnetic Moment**:
The magnetic moment \( M \) of a magnetized wire is given by the formula:
\[ M = m \cdot L \]
where \( m \) is the pole strength and \( L \) is the length of the wire.

2. **Initial Setup**:
We are given that the magnetic moment of the straight wire is \( M \). Therefore, we can express this as:
\[ M = m \cdot L \]

3. **Bending the Wire**:
When the wire is bent into a semicircular arc, the distance between the north and south poles (which are at the ends of the wire) changes. The length of the wire remains the same, but the configuration changes.

4. **Length of the Semicircular Arc**:
The length of the wire remains \( L \), and when bent into a semicircle, the relationship between the radius \( R \) of the semicircle and the length \( L \) is:
\[ L = \pi R \]
Therefore, we can express the radius \( R \) as:
\[ R = \frac{L}{\pi} \]

5. **New Magnetic Moment Calculation**:
In the semicircular configuration, the effective length between the poles is the diameter of the semicircle, which is \( 2R \). Thus, the new magnetic moment \( M' \) can be calculated as:
\[ M' = m \cdot (2R) \]
Substituting \( R \) from the previous step:
\[ M' = m \cdot (2 \cdot \frac{L}{\pi}) = \frac{2mL}{\pi} \]

6. **Substituting for \( mL \)**:
Recall that \( mL = M \). Therefore, we can substitute \( M \) into the equation:
\[ M' = \frac{2M}{\pi} \]

7. **Final Result**:
The new magnetic moment when the wire is bent into a semicircular arc is:
\[ M' = \frac{2M}{\pi} \]

### Conclusion:
The magnetic moment of the magnetized steel wire when bent to form a semicircular arc becomes \( \frac{2M}{\pi} \).