Question

A long solenoid of radius R and length L has n turns per unit length. A circular loop of radius r(<R) is placed inside at the centre of the solenoid such that its axis coincides with the axis of the solenoid. Obtain the mutual inductance of the solenoid and the loop.

Hint:

Notice that the mutual inductance depends strictly on the area of the smaller inner loop ($\pi r^2$), not the larger outer solenoid. The magnetic field only links flux physically through the area where the secondary loop actually exists.

Answer 1

Concept:
• To calculate the mutual inductance $M$ rigidly existing between any two inductive geometries, we pass a theoretical current $I$ through the primary structure to find its uniform magnetic field.
• Next, we mathematically compute the total magnetic flux $\Phi$ actively linking the secondary structure due exclusively to this generated field.
• The mutual inductance is then securely extracted using the fundamental proportionality equation: $M = \frac{\Phi}{I}$.

Step 1: Determine the magnetic field of the primary solenoid
Let us assume an active current $I$ flows continuously through the long outer solenoid.
The solenoid aggressively generates a largely uniform magnetic field strictly along its central axis, penetrating its hollow interior.
The magnitude of this magnetic field $B$ is universally given by Ampere's Law formulation:
\[ B = \mu_0 n I \]
where $\mu_0$ is the permeability of free space and $n$ represents the number of turns strictly per unit length.

Step 2: Calculate the magnetic flux linked with the inner loop
The smaller circular loop of radius $r$ is situated directly inside the solenoid, sharing the exact same central axis.
Because their geometric axes completely coincide, the generated magnetic field $B$ is strictly perpendicular to the flat area $A$ of the inner loop.
The total cross-sectional area of the small inner loop is mathematically exactly:
\[ A = \pi r^2 \]
The total magnetic flux $\Phi$ successfully penetrating through this single circular loop is the product of the field and its area:
\[ \Phi = B \cdot A = B A \cos(0^\circ) \]
Substitute the previously derived expression for the magnetic field:
\[ \Phi = (\mu_0 n I) \times (\pi r^2) \]
\[ \Phi = \mu_0 n \pi r^2 I \]

Step 3: Extract the expression for Mutual Inductance
By fundamental definition, the mutual inductance $M$ tightly relates the linked flux to the driving source current:
\[ \Phi = M I \]
Comparing our geometrically derived flux expression directly with this definitional formula strictly isolates $M$:
\[ M = \frac{\Phi}{I} = \frac{\mu_0 n \pi r^2 I}{I} \]
The driving current $I$ perfectly cancels out, leaving the final physical geometry-dependent expression:
\[ M = \mu_0 n \pi r^2 \]

Step 4: Conclusion
The formal expression for the mutual inductance seamlessly existing between the long solenoid and the tiny coaxial inner loop is solidly exactly $\mu_0 n \pi r^2$.