(C) Let the length of the wire be $L$. For a coil with $N$ turns and radius $R_1$, the circumference is $2\pi R_1 = L/N$, so $R_1 = L/(2\pi N)$. The m

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(C) Let the length of the wire be $L$. For a coil with $N$ turns and radius $R_1$, the circumference is $2\pi R_1 = L/N$, so $R_1 = L/(2\pi N)$. The magnetic field at the center is $B_1 = \frac{\mu_0 N I}{2 R_1} = \frac{\mu_0 N I}{2 (L / 2\pi N)} = \frac{\mu_0 \pi N^2 I}{L}$. Similarly, for a coil with $n$ turns and radius $R_2$, the magnetic field is $B_2 = \frac{\mu_0 n^2 \pi I}{L}$. The ratio of the magnetic fields is $B_1 / B_2 = (\frac{\mu_0 \pi N^2 I}{L}) / (\frac{\mu_0 \pi n^2 I}{L}) = N^2 / n^2$.

Class 12 Physics Moving Charges and Magnetism Biot-Savart's Law and its application

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$A$ long conducting wire having a current $I$ flowing through it is bent into a circular coil of $N$ turns. Then it is bent into a circular coil of $n$ turns. The magnetic field is calculated at the centre of the coils in both cases. The ratio of the magnetic field in the first case to that of the second case is:

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A
$N : n$
B
$n^2 : N^2$
C
$N^2 : n^2$
D
$n : N$

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Class 12

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Moving Charges and Magnetism

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Biot-Savart's Law and its application

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$A$ square frame of side $l$ carries a current $i$. The magnetic field at its centre is $B$. The same current is passed through a circular coil having the same perimeter as the square. The field at the centre of the circular coil is $B^{\prime}$. The ratio of $\frac{B}{B^{\prime}}$ is

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