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\(^2\). \[ \text{(Given } \pi = \frac{22}{7} \text{)} \]
Correct Answer: 28
Approach Solution - 1
The magnetic moment of a bar magnet is defined as the product of its pole strength \(p\) and the distance \(d\) between the poles. For a straight magnetic strip with a magnetic moment \(M=44 \, \text{A m}^2\), it has a linear shape.
When the strip is bent into a semicircular shape, the effective distance between the poles changes. The length of the strip remains constant. Consider the original distance as \(d\), then \(M=p \cdot d\). Now, for the semicircular shape, the distance between the poles becomes the diameter of the semicircle.
The length of the strip (originally the distance between poles) becomes the perimeter of the semicircle: \(d = \pi r\). Thus, the new distance \(d'\) is twice the radius (\(2r\)), which is also the diameter of the semicircle: \(d' = 2r\).
Setting the two distances equal gives \(d = \pi r = 2r\), resulting in \(r = \frac{d}{\pi}\). Therefore, \(d' = 2 \times \frac{d}{\pi} = \frac{2d}{\pi}\).
To find the new magnetic moment \(M'\):
\[ M' = p \cdot d' = p \cdot \frac{2d}{\pi} = \frac{2}{\pi}(p \cdot d) \]
Substituting the original magnetic moment:
\[ M' = \frac{2}{\pi} \times 44 = \frac{88}{\pi} \]
Given \(\pi = \frac{22}{7}\), calculate:
\[ M' = \frac{88}{\frac{22}{7}} = 88 \times \frac{7}{22} = 4 \times 7 = 28 \, \text{A m}^2 \]
The new magnetic moment is \(28 \, \text{A m}^2\), which falls within the given range \(28, 28\).
Approach Solution -2
Given: The magnetic moment of the straight strip is $44 \, \mathrm{Am}^2$. The strip is bent into a semicircular shape.
When a straight magnetic strip is bent into a semicircular shape, the magnetic moment changes based on the geometry. The formula for the magnetic moment $M$ of a circular loop is given by:
\[ M = I \times A, \]
where $I$ is the current and $A$ is the area of the loop. For a semicircular loop, the area is half of the area of a full circle.
The magnetic moment $M'$ after bending is:
\[ M' = \frac{M}{2}. \]
So, the magnetic moment after bending the strip into a semicircular shape is:
\[ M' = \frac{44}{2} = 28 \, \mathrm{Am}^2. \]
Thus, the correct answer is 28.
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