One of the two identical conducting wires of length L is bent in the form of a circular loop and the other one into a circular coil of N identical turns. If the same current is passed in both, the radio of the magnetic field at the central of the loop (BL) to that at the center of the coil (BC), i.e. B_L \over B_C will be :

Magnetics PYQ — JEE Main 2019 · 9 Jan · Shift 2 · Q58 with Solution

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Correct answer: D

If a wire of length $L$ is bent into a single circular loop of radius $R_1$ , then $L = 2\pi R_1 \implies R_1 = \frac{L}{2\pi}.$

The magnetic field at the center of a single circular loop carrying current $I$ is $B_L = \frac{\mu_0 I}{2R_1}.$

Substitute $R_1 = \frac{L}{2\pi}$ : $B_L = \frac{\mu_0 I}{2\left(\frac{L}{2\pi}\right)} = \frac{\mu_0 I\pi}{L}.$

Now the second identical wire of the same length $L$ is bent into a coil of $N$ identical turns.

If radius of each turn is $R_2$ , then total wire length is $L = N(2\pi R_2).$ So, $R_2 = \frac{L}{2\pi N}.$

The magnetic field at the center of an $N$ -turn coil is $B_C = \frac{\mu_0 N I}{2R_2}.$

Substitute $R_2 = \frac{L}{2\pi N}$ :

= \frac{\mu_0 N I}{\frac{L}{\pi N}} = \frac{\mu_0 \pi N^2 I}{L}.$$ 3. **Find the ratio** $$\frac{B_L}{B_C} = \frac{\frac{\mu_0 I\pi}{L}}{\frac{\mu_0 \pi N^2 I}{L}} = \frac{1}{N^2}.$$ 4. **Option check** - A: $N$ ❌ - B: $\frac{1}{N}$ ❌ - C: $N^2$ ❌ - D: $\frac{1}{N^2}$ ✅ Hence, $$\boxed{\frac{B_L}{B_C} = \frac{1}{N^2}}.$$