The magnetic field at the centre of a current | vedclass

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(B) The magnetic field at the centre of a circular coil of radius $r$ carrying current $I$ is given by $B = \frac{\mu_0 I}{2r}$.Given the area $A = \p

Vedclass · Accepted Answer

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

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(B) The magnetic field at the centre of a circular coil of radius $r$ carrying current $I$ is given by $B = \frac{\mu_0 I}{2r}$.Given the area $A = \pi r^2$, we have $r = \sqrt{\frac{A}{\pi}}$.Substituting $r$ into the magnetic field formula: $B = \frac{\mu_0 I}{2 \sqrt{A/\pi}} = \frac{\mu_0 I \sqrt{\pi}}{2 \sqrt{A}}$.Solving for current $I$: $I = \frac{2 B \sqrt{A}}{\mu_0 \sqrt{\pi}}$.The magnetic moment $M$ is given by $M = I A$.Substituting the value of $I$: $M = \left( \frac{2 B \sqrt{A}}{\mu_0 \sqrt{\pi}} \right) A = \frac{2 B A^{3/2}}{\mu_0 \sqrt{\pi}}$.

The magnetic field at the centre of a current carrying circular coil of area $A$ is $B$. The magnetic moment $(M)$ of the coil is $(\mu_0 = \text{permeability of free space})$

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