The electric current in a circular coil of four turns produces a magnetic induction 32 T at its centre. The coil is unwound and is rewound into a circular coil of single turn, the magnetic induction at the centre of the coil by the same current will be :

Magnetics PYQ — JEE Main 2023 · 29 Jan · Shift 2 · Q57 with Solution

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Correct answer: A

For a coil of $N$ turns and radius $R$ , carrying current $I$ , the magnetic field at the centre is

B = \frac{\mu_0 N I}{2R}

Initially, the coil has:

So,

B_1 = \frac{\mu_0 N_1 I}{2R_1} = \frac{\mu_0 (4) I}{2R_1}

The wire length remains the same.

If the initial 4-turn coil has radius $R_1$ , total wire length is

L = 4(2\pi R_1) = 8\pi R_1

After rewinding into a single turn of radius $R_2$ ,

L = 2\pi R_2

Equating both:

2\pi R_2 = 8\pi R_1

R_2 = 4R_1

Now $N_2 = 1$ and $R_2 = 4R_1$ .

Thus,

B_2 = \frac{\mu_0 N_2 I}{2R_2} = \frac{\mu_0 I}{2(4R_1)} = \frac{\mu_0 I}{8R_1}

Compare with initial field:

B_1 = \frac{4\mu_0 I}{2R_1} = \frac{2\mu_0 I}{R_1}

Therefore,

\frac{B_2}{B_1} = \frac{\mu_0 I/(8R_1)}{2\mu_0 I/R_1} = \frac{1}{16}

So,

B_2 = \frac{B_1}{16} = \frac{32}{16} = 2\,\text{T}

Hence, the correct answer is A.