Q133 / 150 · MHT CET 2026 (15 April Shift 2) · Physics
A long wire is bent into a circular coil of one turn and then into a circular coil of smaller radius having n turns. If the same current is passed in both the cases, the ratio of magnetic field produced at the centre for one turn to that of n turns is
- A$1 : n$
- B$n : 1$
- Correct answerC$1 : n^2$
- D$n^2 : 1$
Solution
Let the length of the wire be $L$.
For a circular coil of one turn, the radius $R_1$ is given by $L = 2 \pi R_1 \Rightarrow R_1 = \dfrac{L}{2 \pi}$.
The magnetic field at the centre of the coil of one turn is $B_1 = \dfrac{\mu_0 I}{2 R_1}$.
For a circular coil of $n$ turns, the radius $R_n$ is given by $L = n(2 \pi R_n) \Rightarrow R_n = \dfrac{L}{2 \pi n} = \dfrac{R_1}{n}$.
The magnetic field at the centre of the coil of $n$ turns is $B_n = \dfrac{\mu_0 n I}{2 R_n}$.
Substituting $R_n = \dfrac{R_1}{n}$, we get $B_n = \dfrac{\mu_0 n I}{2 \left(\dfrac{R_1}{n}\right)} = n^2 \left(\dfrac{\mu_0 I}{2 R_1}\right) = n^2 B_1$.
The ratio of the magnetic field for one turn to that of $n$ turns is $\dfrac{B_1}{B_n} = \dfrac{1}{n^2}$.
Answer: $1 : n^2$
For a circular coil of one turn, the radius $R_1$ is given by $L = 2 \pi R_1 \Rightarrow R_1 = \dfrac{L}{2 \pi}$.
The magnetic field at the centre of the coil of one turn is $B_1 = \dfrac{\mu_0 I}{2 R_1}$.
For a circular coil of $n$ turns, the radius $R_n$ is given by $L = n(2 \pi R_n) \Rightarrow R_n = \dfrac{L}{2 \pi n} = \dfrac{R_1}{n}$.
The magnetic field at the centre of the coil of $n$ turns is $B_n = \dfrac{\mu_0 n I}{2 R_n}$.
Substituting $R_n = \dfrac{R_1}{n}$, we get $B_n = \dfrac{\mu_0 n I}{2 \left(\dfrac{R_1}{n}\right)} = n^2 \left(\dfrac{\mu_0 I}{2 R_1}\right) = n^2 B_1$.
The ratio of the magnetic field for one turn to that of $n$ turns is $\dfrac{B_1}{B_n} = \dfrac{1}{n^2}$.
Answer: $1 : n^2$
Hence, the correct answer is option: 'C'.