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(A) The magnetic dipole moment $M$ of a current loop is given by $M = I A$,where $I$ is the current and $A$ is the area of the loop.For a circular loo

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$\frac{2 \pi B r^3}{\mu_0}$

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(A) The magnetic dipole moment $M$ of a current loop is given by $M = I A$,where $I$ is the current and $A$ is the area of the loop.For a circular loop of radius $r$,the area is $A = \pi r^2$.Thus,$M = I (\pi r^2) \quad \dots(1)$The magnetic field $B$ at the centre of a circular loop carrying current $I$ is given by $B = \frac{\mu_0 I}{2r}$.Rearranging this formula to solve for current $I$,we get $I = \frac{2rB}{\mu_0}$.Substituting the expression for $I$ into equation $(1)$:$M = \left( \frac{2rB}{\mu_0} \right) (\pi r^2)$$M = \frac{2 \pi B r^3}{\mu_0}$.

$A$ circular loop having radius $r$,carrying current $I$,produces a magnetic field at the centre of the loop equal to $B$. What will be the magnetic dipole moment of this loop?

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