1. Magnetic moment of a bar magnet

For a magnetic strip (bar magnet), magnetic moment is $M = m \times 2l$ where:

• $m$ = pole strength
• $2l$ = magnetic length

Given initially, $M_0 = 44\,\text{A m}^2$

So, $m \cdot 2l = 44$

2. When bent into a semicircle

The strip length remains unchanged.

Originally, its length is $2l$. When bent into a semicircle of radius $r$, the arc length is $\pi r = 2l$

Hence, $r = \frac{2l}{\pi}$

3. New magnetic moment

After bending into a semicircle, the pole strength $m$ remains same, but the effective magnetic length becomes the straight-line distance between the poles, i.e. the diameter of the semicircle: $\text{new magnetic length} = 2r$

Therefore new magnetic moment, $M' = m \cdot 2r$

Using $r = \frac{2l}{\pi}$, $M' = m \cdot 2\left(\frac{2l}{\pi}\right) = \frac{2m(2l)}{\pi}$

But $m(2l) = 44$, so $M' = \frac{2 \times 44}{\pi} = \frac{88}{\pi}$

Given $\pi = \frac{22}{7}$

Thus, $M' = \frac{88}{22/7} = 88 \times \frac{7}{22} = 4 \times 7 = 28$

4. Final Answer

$\boxed{28\,\text{A m}^2}$