Question

A straight magnetic strip has a magnetic moment of \( 44 \, \text{A m}^2 \). If the strip is bent in a semicircular shape, its magnetic moment will be \( \dots \) A m\(^2\). \[ \text{(Given } \pi = \frac{22}{7} \text{)} \]

Answer 1

Correct Answer: 28

The magnetic moment \(M\) of a bar magnet is defined as the product of its pole strength \(p\) and the distance \(d\) between the poles, such that \(M=p \cdot d\). For a straight magnetic strip with a magnetic moment \(M=44 \, \text{A m}^2\), its shape is linear.

When this strip is bent into a semicircular shape, the distance between the poles changes. The strip's length, however, remains constant. Let the original distance between poles be \(d\). In the semicircular configuration, the distance between the poles becomes the diameter of the semicircle.

The original distance between poles, which is the length of the strip, now forms the perimeter of the semicircle: \(d = \pi r\). The new distance between the poles, \(d'\), is the diameter of this semicircle, which is twice the radius: \(d' = 2r\).

Equating the original distance \(d\) to the perimeter of the semicircle gives \(d = \pi r\). From this, we find the radius \(r = \frac{d}{\pi}\). Consequently, the new distance between the poles is \(d' = 2r = 2 \times \frac{d}{\pi} = \frac{2d}{\pi}\).

The new magnetic moment \(M'\) can be calculated as:

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

Substituting the original magnetic moment \(M = p \cdot d = 44 \, \text{A m}^2\):

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

Using the approximation \(\pi = \frac{22}{7}\), we calculate \(M'\):

\[ M' = \frac{88}{\frac{22}{7}} = 88 \times \frac{7}{22} = 4 \times 7 = 28 \, \text{A m}^2 \]

The resulting new magnetic moment is \(28 \, \text{A m}^2\), which falls within the specified range of \(28, 28\).