JEE Physics 2023 Question with Solution

Given: A long conducting wire carrying current $I$ is first bent into a circular coil of $N$ turns and then into a circular coil of $n$ turns.

Find: The ratio of magnetic field at the centre in the first case to that in the second case.

For a circular coil,

$$B = \frac{\mu_0 I N}{2r}$$

Also, if the same wire is used, its total length remains constant. Hence,

$$L = 2\pi r N$$

So,

$$r \propto \frac{1}{N}$$

Therefore,

$$B \propto \frac{N}{r} \propto N^2$$

Thus, for the two cases,

$$\frac{B_1}{B_2} = \frac{N^2}{n^2}$$

Therefore, the correct option is C, that is $N^2 : n^2$.

Given: The same conducting wire is bent into coils of $N$ turns and $n$ turns.

Find: $\frac{B_1}{B_2}$.

From the extracted working,

$$L = (2\pi r)n$$

Hence,

$$r \propto \frac{L}{n}$$

So the magnetic field at the centre becomes

$$B = n\left(\frac{\mu_0 i}{2r}\right) \propto n\left(\frac{1}{r}\right)$$

Using $r \propto \frac{L}{n}$,

$$B \propto n\left(\frac{n}{L}\right) = \frac{n^2}{L}$$

Since the wire length $L$ and current remain the same, magnetic field is proportional to the square of the number of turns. Thus,

$$\frac{B_1}{B_2} = \frac{N^2}{n^2}$$

So, the correct answer is $N^2 : n^2$.