If a wire of length L forms a loop of radius | vedclass

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(C) The total length of the wire $L$ is equal to the circumference of the loop multiplied by the number of turns $n$: $L = n(2\pi R)$.From this,we fin

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$\frac{\mu_0 I}{4\pi} \frac{4\pi^2 n^2}{L}$

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(C) The total length of the wire $L$ is equal to the circumference of the loop multiplied by the number of turns $n$: $L = n(2\pi R)$.From this,we find the radius $R = \frac{L}{2\pi n}$.The magnetic field $B$ at the center of a circular coil with $n$ turns is given by $B = n \left( \frac{\mu_0 I}{2R} \right)$.Substituting the value of $R$ into the formula: $B = n \left( \frac{\mu_0 I}{2(L / 2\pi n)} \right)$.Simplifying the expression: $B = n \left( \frac{\mu_0 I \cdot 2\pi n}{2L} \right) = \frac{\mu_0 I \cdot 2\pi n^2}{2L} = \frac{\mu_0 I \cdot \pi n^2}{L}$.To match the given options,we multiply and divide by $4\pi$: $B = \frac{\mu_0 I}{4\pi} \cdot \frac{4\pi^2 n^2}{L}$.

$(i)$	$(ii)$	$(iii)$
$(A). \frac{\mu_0 i}{2r} \odot$	$(A). \frac{\mu_0}{2\pi} \frac{i}{r}(\pi - 2)$	$(A). \frac{\mu_0}{2r} \frac{2i}{r}(\pi + 1) \otimes$
$(B). \frac{\mu_0 i}{2r} \otimes$	$(B). \frac{\mu_0 i}{4\pi} \frac{i}{r}(\pi + 2) \otimes$	$(B). \frac{\mu_0 i}{4r} \frac{2i}{r}(\pi - 1) \otimes$
$(C). \frac{3\mu_0 i}{8r} \otimes$	$(C). \frac{\mu_0 i}{4r} \otimes$	$(C). \text{Zero}$
$(D). \frac{3\mu_0 i}{8r} \odot$	$(D). \frac{\mu_0 i}{4r} \odot$	$(D). \text{Infinite}$

If a wire of length $L$ forms a loop of radius $R$ and has $n$ turns,find the magnetic field at the center of the loop if the current flowing in the loop is $I$.

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