Question

A long solenoid of length \(L\) and radius \(r_1\) having \(N_1\) turns is surrounded symmetrically by a coil of radius \(r_2\,(>r_1)\) having \(N_2\) turns \((N_2 \ll N_1)\) around its mid-point. Derive an expression for the mutual inductance of the solenoid and the coil. Is \(M_{12}=M_{21}\) valid in this case?

Hint:

For a long solenoid, \[ B=\mu_0 n I. \] When calculating mutual inductance, always use only the region where magnetic field actually exists. In this problem, the effective area is \(\pi r_1^2\), not \(\pi r_2^2\).

Answer 1

Concept: Mutual inductance between two coils is defined as the magnetic flux linked with one coil due to current flowing in the other coil divided by that current. Mathematically, \[ M=\frac{N\Phi}{I}. \] For a long solenoid, the magnetic field inside the solenoid is uniform and is given by \[ B=\mu_0 n I, \] where \[ n=\frac{N_1}{L} \] is the number of turns per unit length. Hence, \[ B=\mu_0\frac{N_1}{L}I_1. \] Since the field outside a long solenoid is negligible, only the area of the solenoid contributes to the flux linkage.

Step 1: Calculate the magnetic field produced by the solenoid. Let a current \(I_1\) flow through the long solenoid. For a long solenoid, \[ B=\mu_0\frac{N_1}{L}I_1. \] This magnetic field exists only inside the solenoid.

Step 2: Calculate the magnetic flux linked with one turn of the surrounding coil. The surrounding coil has radius \(r_2\), but the magnetic field exists only inside the solenoid of radius \(r_1\). Therefore, effective area through which flux passes is \[ A=\pi r_1^2. \] Hence flux through one turn of the outer coil is \[ \Phi = BA. \] Substituting the value of \(B\), \[ \Phi = \left( \mu_0\frac{N_1}{L}I_1 \right) \pi r_1^2. \] Therefore, \[ \Phi = \mu_0\frac{N_1}{L}I_1\pi r_1^2. \]

Step 3: Calculate total flux linkage with the outer coil. The outer coil contains \(N_2\) turns. Therefore total flux linkage is \[ N_2\Phi = N_2 \left( \mu_0\frac{N_1}{L}I_1\pi r_1^2 \right). \] Hence, \[ N_2\Phi = \mu_0\frac{N_1N_2}{L}\pi r_1^2 I_1. \]

Step 4: Use the definition of mutual inductance. By definition, \[ M = \frac{N_2\Phi}{I_1}. \] Substituting the above expression, \[ M = \frac{ \mu_0\frac{N_1N_2}{L}\pi r_1^2 I_1 } {I_1}. \] Therefore, \[ \boxed{ M = \mu_0 \frac{N_1N_2\pi r_1^2}{L} }. \] This is the required expression for mutual inductance.

Step 5: Discuss whether \(M_{12}=M_{21}\). According to the reciprocity theorem of mutual induction, \[ M_{12}=M_{21}. \] This result is independent of the sizes and shapes of the two circuits. Therefore, even in the present case, \[ \boxed{M_{12}=M_{21}}. \] Final Answer: \[ \boxed{ M = \mu_0 \frac{N_1N_2\pi r_1^2}{L} } \] and \[ \boxed{ M_{12}=M_{21}. } \]