A straight magnetic strip has a magnetic moment of 44 A m^2. If the strip is bent in a semicircular shape, its magnetic

Question

A straight magnetic strip has a magnetic moment of 44 A m^2. If the strip is bent in a semicircular shape, its magnetic moment will be:

Accepted Answer

Given: The magnetic moment of the straight strip is $M = 44 \, \text{A m}^2$ .

Find: The magnetic moment when the strip is bent into a semicircular shape.

The magnetic moment of a bar magnet is defined as the product of pole strength and magnetic length:

M = p d

where $p$ is the pole strength and $d$ is the distance between the poles.

When the strip is bent into a semicircle, the length of the strip remains constant. If the original magnetic length is $d$ , then for the semicircle:

d = \pi r

The new distance between the poles becomes the diameter:

d' = 2r

From

d = \pi r

we get

r = \frac{d}{\pi}

So,

d' = 2r = \frac{2d}{\pi}

Hence the new magnetic moment is

M' = p d' = p \cdot \frac{2d}{\pi} = \frac{2}{\pi}(p d)

Therefore,

M' = \frac{2}{\pi} M = \frac{2}{\pi} \times 44 = \frac{88}{\pi}

Using

\pi = \frac{22}{7}

we get

M' = \frac{88}{22/7} = 88 \times \frac{7}{22} = 28 \, \text{A m}^2

Therefore, the new magnetic moment is $28 \, \text{A m}^2$ . The correct option is C.

The solution explicitly concludes that the new magnetic moment is $28 \, \text{A m}^2$ . This matches option C.

The second approach shown on the page also ends with $28 \, \text{A m}^2$ , although its intermediate reasoning is not consistent with the standard bar-magnet definition. The first approach provides the defensible derivation, so the correct answer is C.

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