Question

Two concentric circular coils of \(n\) turns each are situated in the same plane. Their radii are \(a_1\) and \(a_2\) (\(a_2>a_1\)) and they carry currents \(I_1\) and \(I_2\) respectively (\(I_1>I_2\)) in opposite direction. The magnetic field at the centre is

• A. \( \dfrac{\mu_0 n}{2}\left[\dfrac{I_1 a_2 - I_2 a_1}{a_1 a_2}\right] \)
• B. \( \dfrac{\mu_0 n}{2 a_1 a_2}(I_1 - I_2) \)
• C. \( \dfrac{\mu_0 n}{2 I_1 I_2}(a_2 - a_1) \)
• D. \( \dfrac{\mu_0 n}{2}\left[\dfrac{I_1 a_1 - I_2 a_2}{a_1 a_2}\right] \)

Hint:

For concentric coils, calculate magnetic fields separately and add algebraically considering direction.

Answer 1

The Correct Option is A

Step 1: Magnetic field due to a circular coil.
Magnetic field at the centre of a circular coil of radius \(a\) carrying current \(I\) is \[ B = \frac{\mu_0 n I}{2a}. \]
Step 2: Fields due to individual coils.
For inner coil: \[ B_1 = \frac{\mu_0 n I_1}{2a_1}. \] For outer coil: \[ B_2 = \frac{\mu_0 n I_2}{2a_2}. \]
Step 3: Net magnetic field.
Since currents are in opposite directions, the resultant field is \[ B = B_1 - B_2 = \frac{\mu_0 n}{2}\left(\frac{I_1}{a_1} - \frac{I_2}{a_2}\right). \] \[ B = \frac{\mu_0 n}{2}\left[\frac{I_1 a_2 - I_2 a_1}{a_1 a_2}\right]. \]
Step 4: Conclusion.
The magnetic field at the centre is given by option (A).