Question

A long conducting wire having a current I flowing through it, is bent into a circular coil of $N$ turns .Then it is bent into a circular coil of $n$ turns. The magnetic field is calculated at the centre of coils in both the cases. The ratio of the magnetic field in first case to that of second case is :

• A. $n : N$
• B. $N : n$
• C. $N ^2: n ^2$
• D. $n^2: N ^2$

Hint:

The magnetic field at the centre of a coil is directly proportional to the square of the number of turns in the coil.

Answer 1

The Correct Option is C

\(I=(2\pi\ r)n\)
\(r∝(\frac{I}{n})\)
\(B=n(\frac{\mu_0i}{2r})∝(\frac{\mu_0i}{2L})n^2\)
\(\frac{B_1}{B_2}=(\frac{N^2}{n^2})\)
So, the correct answer is (C) : \(N^2:n^2\)

Answer 2

The magnetic field at the centre of a circular coil is given by the formula: \[ B = \frac{\mu_0 I N}{2r} \] where \( \mu_0 \) is the permeability of free space, \( I \) is the current, \( N \) is the number of turns, and \( r \) is the radius of the coil. Since the current and the radius are constant, the ratio of the magnetic field in the two cases is: \[ \frac{B_1}{B_2} = \frac{N_1^2}{N_2^2} = \frac{N^2}{n^2} \]