A wire in the form of a circular loop of one | vedclass

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Question

(C) The magnetic field at the center of a circular coil of $n$ turns carrying current $I$ and having radius $r$ is given by $B = \frac{\mu_0 n I}{2r}$

Vedclass · Accepted Answer

$4B$

Vedclass · Answer

(C) The magnetic field at the center of a circular coil of $n$ turns carrying current $I$ and having radius $r$ is given by $B = \frac{\mu_0 n I}{2r}$.If the same wire of length $L$ is used,then for $n=1$,$L = 2\pi r_1$,so $r_1 = \frac{L}{2\pi}$. Thus,$B = \frac{\mu_0 I}{2(L/2\pi)} = \frac{\mu_0 I \pi}{L}$.For $n=2$,the new radius $r_2$ is given by $L = 2(2\pi r_2)$,so $r_2 = \frac{L}{4\pi} = \frac{r_1}{2}$.The new magnetic field $B'$ is $B' = \frac{\mu_0 (2) I}{2r_2} = \frac{\mu_0 I}{r_2} = \frac{\mu_0 I}{r_1/2} = 2 \left( \frac{\mu_0 I}{r_1} \right) = 4 \left( \frac{\mu_0 I}{2r_1} \right) = 4B$.Therefore,the new magnetic field is $4B$.

$A$ wire in the form of a circular loop of one turn carrying a current produces a magnetic field $B$ at the centre. If the same wire is looped into a coil of two turns and carries the same current,the new value of magnetic induction at the centre is

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