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:

Solution :

Magnetic field at the centre of a circular coil is given by :

$$B = \frac{\mu_0 NI}{2R}$$

Given initial number of turns :

$$N_1 = 4$$

Initial magnetic field :

$$B_1 = 32\text{ T}$$

When the wire is rewound into a single turn coil :

$$N_2 = 1$$

Since length of wire remains constant :

$$2\pi R_1N_1 = 2\pi R_2N_2$$

$$R_2 = 4R_1$$

Now,

$$B_2 = \frac{\mu_0 N_2 I}{2R_2}$$

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

$$= \frac{\mu_0 I}{8R_1}$$

Initial field :

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

$$= \frac{2\mu_0 I}{R_1}$$

Therefore,

$$\frac{B_2}{B_1} = \frac{\mu_0 I/8R_1}{2\mu_0 I/R_1}$$

$$= \frac{1}{16}$$

Hence,

$$B_2 = \frac{32}{16}$$

$$= 2\text{ T}$$

Final Answer :

$$2\text{ T}$$