Solution:
For circle: \(M = I \pi R^2\) where \(2\pi R = L ⇒ R = \frac{L}{2\pi}\), so \(M = \frac{I L^2}{4\pi}\). For square of side \(a = \frac{L}{4}\): \(M' = I a^2 = \frac{I L^2}{16}\). Thus, \(M' = \frac{\pi}{4} M\).
A thin circular wire carrying a current \(I\) has a magnetic moment \(M\). The shape of the wire is changed to a square and it carries the same current. It will have a magnetic moment of:
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