A bar magnet of magnetic moment M is bent in the form of quadrant of circular arc . The new magnetic moment is

Options

A. `(2M)/(pi)`
B. `(2 sqrt2 M)/(pi)`
C. `(sqrt2M)/(pi)`
D. M

Answer

To solve the problem of finding the new magnetic moment of a bar magnet bent into the form of a quadrant of a circular arc, we can follow these steps:

### Step-by-Step Solution:

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

2. **Visualize the Bending of the Magnet**:
When the bar magnet is bent into the shape of a quadrant of a circular arc, the effective length of the magnet changes. The new effective length will be the straight line distance between the north and south poles of the arc.

3. **Determine the Effective Length**:
For a quadrant of a circle with radius \( r \), the effective length (the straight line connecting the two ends of the arc) can be calculated using the Pythagorean theorem:
\[ \text{Effective Length} = \sqrt{(r)^2 + (r)^2} = \sqrt{2r^2} = r\sqrt{2} \]

4. **Relate the Radius to the Original Length**:
The length of the arc (which is the original length \( L \) of the bar magnet) is given by the formula for the circumference of a circle:
\[ \text{Arc Length} = \frac{\pi r}{2} \]
Setting this equal to \( L \):
\[ \frac{\pi r}{2} = L \implies r = \frac{2L}{\pi} \]

5. **Substitute the Radius into the Effective Length**:
Now substitute \( r \) back into the effective length:
\[ \text{Effective Length} = r\sqrt{2} = \left(\frac{2L}{\pi}\right)\sqrt{2} = \frac{2L\sqrt{2}}{\pi} \]

6. **Calculate the New Magnetic Moment**:
The new magnetic moment \( M' \) can now be calculated using the new effective length:
\[ M' = m \cdot \text{Effective Length} = m \cdot \left(\frac{2L\sqrt{2}}{\pi}\right) \]
Since the original magnetic moment \( M = mL \), we can express \( m \) as:
\[ m = \frac{M}{L} \]
Substituting this into the equation for \( M' \):
\[ M' = \frac{M}{L} \cdot \left(\frac{2L\sqrt{2}}{\pi}\right) = \frac{2\sqrt{2}M}{\pi} \]

### Final Result:
Thus, the new magnetic moment \( M' \) when the bar magnet is bent into the form of a quadrant of a circular arc is:
\[ M' = \frac{2\sqrt{2}M}{\pi} \]