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

Step 1: Flux through the outer coil due to the solenoid.
Magnetic field exists only inside the solenoid of radius \( r_1 \). Area of solenoid cross-section: \[ A = \pi r_1^2 \] Flux through one turn of the outer coil: \[ \phi = B A = \mu_0 \frac{N_1}{L} I_1 \cdot \pi r_1^2 \]
Step 2: Total flux linkage with the outer coil.
For \( N_2 \) turns in the outer coil: \[ \Phi = N_2 \phi = N_2 \mu_0 \frac{N_1}{L} I_1 \pi r_1^2 \]
Step 3: Calculate mutual inductance. \[ M_{12} = \frac{\Phi}{I_1} = \mu_0 \frac{N_1 N_2}{L} \pi r_1^2 \]
Step 4: Symmetry of mutual inductance. \[ M_{12} = M_{21} \] Mutual inductance is symmetric, independent of the coil geometry, as long as the medium is linear and isotropic.
Step 5: Final Expression. \[ M = \mu_0 \frac{N_1 N_2}{L} \pi r_1^2 \] Yes, \( M_{12} = M_{21} \) holds true.

Answer 2

Another way to reach the mutual inductance is through the vector potential \( \mathbf{A} \) of the solenoid, using \( \Phi = \oint \mathbf{A} \cdot d\mathbf{l} \) instead of directly multiplying field by area.

Step 1: Vector potential of a long solenoid.
For a point inside the solenoid, at radial distance \( s \le r_1 \) from the axis, the vector potential circles the axis with magnitude \[A(s) = \frac{\mu_0 N_1 I_1}{2L} \, s\] obtained from \( \oint \mathbf{A} \cdot d\mathbf{l} = \Phi_{\text{enclosed}} \) applied to a circle of radius \( s \), enclosing flux \( B \pi s^2 \). For a point outside the winding (\( s > r_1 \)), all of the solenoid's flux \( B \pi r_1^2 \) is enclosed no matter how far out the circle is drawn, so \[A(s) = \frac{\mu_0 N_1 I_1}{2L} \, \frac{r_1^2}{s}\]

Step 2: Apply this at the outer coil.
The outer coil sits at radius \( r_2 > r_1 \), outside the solenoid winding, so \[A(r_2) = \frac{\mu_0 N_1 I_1}{2L} \, \frac{r_1^2}{r_2}\]

Step 3: Flux linked with one turn of the coil.\[\phi = \oint \mathbf{A} \cdot d\mathbf{l} = A(r_2) \cdot 2\pi r_2 = \frac{\mu_0 N_1 I_1}{2L} \cdot \frac{r_1^2}{r_2} \cdot 2\pi r_2\]The factor of \( r_2 \) cancels exactly:\[\phi = \mu_0 \frac{N_1 I_1}{L} \pi r_1^2\]the same result as before, but now it is clear why the coil's own radius \( r_2 \) drops out: the vector potential outside the solenoid falls off as \( 1/r_2 \) exactly fast enough to cancel the \( 2\pi r_2 \) circumference.

Step 4: Total flux and mutual inductance.
With \( N_2 \) turns on the outer coil,\[\Phi = N_2 \phi = \mu_0 \frac{N_1 N_2}{L} \pi r_1^2 I_1, \qquad M = \frac{\Phi}{I_1} = \mu_0 \frac{N_1 N_2 \pi r_1^2}{L}\]

On whether \( M_{12} = M_{21} \) holds: this identity does not depend on which circuit carries the current, it is a general consequence of the reciprocal nature of magnetic flux linkage between any two fixed loops in space. So yes, \( M_{12} = M_{21} \) is valid for this solenoid-coil pair, exactly as for any two coupled circuits.