A straight magnetic strip has a magnetic moment of $$44 \text{ Am}^2$$. If the strip is bent in a semicircular shape, its magnetic moment will be ______ $$\text{Am}^2$$. (given $$\pi = \frac{22}{7}$$)

We need to find the magnetic moment of a straight magnetic strip after it is bent into a semicircular shape.

Original magnetic moment: $$M = 44 \text{ Am}^2$$

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

For a bar magnet (or magnetic strip), the magnetic moment is defined as:

$$M = m \times L$$

where $$m$$ is the pole strength and $$L$$ is the length of the strip (distance between the poles). Here:

$$44 = m \times L \implies m = \frac{44}{L}$$

When the strip of length $$L$$ is bent into a semicircular shape, the length of the strip equals the semicircular arc:

$$\pi R = L \implies R = \frac{L}{\pi}$$

where $$R$$ is the radius of the semicircle.

In a semicircle, the two ends of the strip (which carry the magnetic poles) are at diametrically opposite points. The distance between them is the diameter:

$$d = 2R = \frac{2L}{\pi}$$

The pole strength $$m$$ remains unchanged (it is an intrinsic property of the magnet). The new magnetic moment is:

$$M' = m \times d = m \times \frac{2L}{\pi} = \frac{2mL}{\pi} = \frac{2M}{\pi}$$

$$M' = \frac{2 \times 44}{\pi} = \frac{88}{\frac{22}{7}} = \frac{88 \times 7}{22} = \frac{616}{22} = 28 \text{ Am}^2$$

The answer is 28 Am$$^2$$.