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Question

(B) The magnetic induction $B$ at the centre of a circular loop with $n$ turns and radius $r$ carrying current $i$ is given by: $B = \frac{\mu_0 n i}{

Vedclass · Accepted Answer

$\frac{0.2 \pi}{\mu_0}$

Vedclass · Answer

(B) The magnetic induction $B$ at the centre of a circular loop with $n$ turns and radius $r$ carrying current $i$ is given by: $B = \frac{\mu_0 n i}{2r}$ From this, we can express the current as: $i = \frac{2 B r}{\mu_0 n}$ The magnetic moment $M$ of the loop is defined as: $M = n i A$ Substituting the expression for $i$: $M = n \left( \frac{2 B r}{\mu_0 n} ight) A = \frac{2 B r A}{\mu_0}$ Given area $A = \pi \ m^2$, we know $A = \pi r^2$, so $\pi r^2 = \pi$, which implies $r = 1 \ m$. Substituting the values $B = 0.1 \ T$, $r = 1 \ m$, and $A = \pi \ m^2$: $M = \frac{2 	imes 0.1 	imes 1 	imes \pi}{\mu_0} = \frac{0.2 \pi}{\mu_0}$

Magnetic induction at the centre of a circular loop of area $\pi \ m^2$ is $0.1 \ T$. The magnetic moment of the loop is ( $\mu_0 = \text{permeability of air}$ ).

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