Two identical wires, carrying equal currents | vedclass

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(A) Let the length of each wire be $L$ and the current flowing through them be $I$. For a coil with $N$ turns and radius $R$, the length of the wire i

Vedclass · Accepted Answer

$4: 9$

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(A) Let the length of each wire be $L$ and the current flowing through them be $I$. For a coil with $N$ turns and radius $R$, the length of the wire is $L = N(2\pi R)$, so $R = L / (2\pi N)$. The magnetic field at the center of a circular coil is given by $B = \frac{\mu_0 N I}{2R}$. Substituting $R$, we get $B = \frac{\mu_0 N I}{2(L / 2\pi N)} = \frac{\mu_0 \pi N^2 I}{L}$. Since $\mu_0$, $\pi$, $I$, and $L$ are constant for both coils, $B \propto N^2$. For coil $A$, $N_A = 2$, so $B_A \propto (2)^2 = 4$. For coil $B$, $N_B = 3$, so $B_B \propto (3)^2 = 9$. The ratio of the magnetic fields is $B_A / B_B = 4 / 9$.

Two identical wires, carrying equal currents are bent into circular coils $A$ and $B$ with $2$ and $3$ turns respectively. The ratio of the magnetic fields at the centres of the coils $A$ and $B$ is

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