Question

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

Hint:

For solenoid–coil systems:

Flux area = cross-section of inner solenoid
Mutual inductance is always symmetric: \( M_{12} = M_{21} \)

Answer 1

Concept: Mutual inductance: \[ M = \frac{\text{Flux linked with secondary}}{\text{Current in primary}} \] Magnetic field inside a long solenoid: \[ B = \mu_0 n I = \mu_0 \frac{N_1}{L} I_1 \] Field is uniform inside solenoid and negligible outside.
Step 1: Flux through outer coil due to solenoid. Magnetic field exists only inside solenoid of radius \( r_1 \). Area contributing to flux: \[ A = \pi r_1^2 \] Flux through one turn of outer coil: \[ \phi = B A = \mu_0 \frac{N_1}{L} I_1 \cdot \pi r_1^2 \]
Step 2: Total flux linkage with outer coil. Outer coil has \( N_2 \) turns: \[ \Phi = N_2 \phi = N_2 \mu_0 \frac{N_1}{L} I_1 \pi r_1^2 \]
Step 3: Mutual inductance. \[ M_{12} = \frac{\Phi}{I_1} \] \[ M_{12} = \mu_0 \frac{N_1 N_2}{L} \pi r_1^2 \]
Step 4: Mutual inductance symmetry. In general: \[ M_{12} = M_{21} \] This is a fundamental property of mutual inductance, independent of geometry (as long as medium is linear and isotropic).
Step 5: Conclusion.

Mutual inductance: \[ M = \mu_0 \frac{N_1 N_2}{L} \pi r_1^2 \]
Yes, \( M_{12} = M_{21} \) is valid.

Answer 2

The magnetic field produced by a solenoid is normally used to find the flux through the outer coil directly. Instead of just stating that mutual inductance is symmetric, it helps to first write the general expression for mutual inductance between two circuits, the Neumann formula, and show that it is inherently symmetric in the two circuits.

Step 1: Neumann's formula for mutual inductance.
For any two current-carrying loops, labelled 1 and 2, the mutual inductance is given by the double line integral:\[M_{12} = \frac{\mu_0}{4\pi} \oint_{C_1} \oint_{C_2} \frac{d\mathbf{l}_1 \cdot d\mathbf{l}_2}{r_{12}}\]where \( r_{12} \) is the distance between a line element \( d\mathbf{l}_1 \) on circuit 1 and a line element \( d\mathbf{l}_2 \) on circuit 2.

Step 2: Notice the built-in symmetry.
Since \( d\mathbf{l}_1 \cdot d\mathbf{l}_2 = d\mathbf{l}_2 \cdot d\mathbf{l}_1 \) and \( r_{12} = r_{21} \), swapping the labels 1 and 2 in the double integral leaves its value completely unchanged:\[M_{12} = \frac{\mu_0}{4\pi} \oint_{C_1} \oint_{C_2} \frac{d\mathbf{l}_1 \cdot d\mathbf{l}_2}{r_{12}} = \frac{\mu_0}{4\pi} \oint_{C_2} \oint_{C_1} \frac{d\mathbf{l}_2 \cdot d\mathbf{l}_1}{r_{21}} = M_{21}\]So \( M_{12} = M_{21} \) is not a coincidence of this particular geometry, it is guaranteed by the mathematical form of the mutual inductance integral itself, true for any pair of loops regardless of shape.

Step 3: Apply this to the solenoid-coil system.
Because \( M_{12} = M_{21} \) always, it is enough to compute whichever one is easier, here that is \( M_{12} \): the flux through the outer coil (radius \( r_2 \)) produced by current \( I_1 \) in the solenoid (radius \( r_1 \), turns \( N_1 \), length \( L \)). Inside a long solenoid the field is uniform, \( B = \mu_0 \frac{N_1}{L} I_1 \), and essentially zero outside the solenoid winding. Since \( r_2 > r_1 \), the only region contributing flux is the solenoid's own cross-section, \( \pi r_1^2 \), not the coil's larger area \( \pi r_2^2 \).

Flux linked with one turn of the outer coil:\[\phi = B \pi r_1^2 = \mu_0 \frac{N_1}{L} I_1 \pi r_1^2\]Total flux linkage with \( N_2 \) turns:\[\Phi = N_2 \phi = \mu_0 \frac{N_1 N_2}{L} \pi r_1^2 I_1\]

Step 4: Mutual inductance.\[M_{12} = \frac{\Phi}{I_1} = \mu_0 \frac{N_1 N_2 \pi r_1^2}{L}\]By the symmetry proved in Step 2, this is also \( M_{21} \), even though computing \( M_{21} \) directly, by finding the field of the wider, loosely wound coil at every point inside the tightly wound solenoid, would have been far harder.

So the mutual inductance of the arrangement is \[M = \mu_0 \frac{N_1 N_2 \pi r_1^2}{L}\] and yes, \( M_{12} = M_{21} \) is valid here, not just as a feature of this particular configuration but as a general result satisfied by every pair of circuits.