The magnetic field at the centre of a current | vedclass

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Question

(A) Let $r$ be the radius of the circular loop. Since the area $A = \pi r^2$,we have $r = \sqrt{\frac{A}{\pi}}$. The magnetic field at the centre of a

Vedclass · Accepted Answer

$\frac{2 B A^{3 / 2}}{\mu_0 \pi^{1 / 2}}$

Vedclass · Answer

(A) Let $r$ be the radius of the circular loop. Since the area $A = \pi r^2$,we have $r = \sqrt{\frac{A}{\pi}}$. The magnetic field at the centre of a circular loop is given by $B = \frac{\mu_0 I}{2r}$. Substituting the value of $r$,we get $B = \frac{\mu_0 I}{2 \sqrt{A/\pi}}$. Solving for current $I$,we get $I = \frac{2B}{\mu_0} \sqrt{\frac{A}{\pi}}$. The magnetic moment $M$ of the loop is defined as $M = I A$. Substituting the expression for $I$,we get $M = \left( \frac{2B}{\mu_0} \sqrt{\frac{A}{\pi}} \right) A = \frac{2 B A^{3/2}}{\mu_0 \sqrt{\pi}}$. Thus,the correct option is $A$.

The magnetic field at the centre of a current carrying circular coil of an area $A$ is $B$. The magnetic moment of the coil is (where $\mu_0$ is the permeability of free space).

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