Q.Two concentric circular coils, one of small radius $r_1$ and the other of large radius $r_2$, such that $r_1 \ll r_2$, are placed co-axially with centres coinciding. Obtain the mutual inductance of the arrangement.

Concept: Mutual Inductance — the flux through one coil due to current in the other.

Step 1: Let current $I$ flow in the larger coil (radius $r_2$). At its centre, the magnetic field is uniform and given by

Step 2: Since $r_1 \ll r_2$, the field over the entire area of the smaller coil is approximately this same $B$. The flux through the smaller coil (area $\pi r_1^2$) is …

Concept and Intuition

Mutual inductance $M$ between two coils is defined by the flux linkage: if a current $I_2$ flows in coil 2, the flux through coil 1 is $\Phi_1 = M I_2$. Equivalently, $M = \frac{\Phi_1}{I_2}$ when coil 1 carries no current. The key is to choose the simpler path: here, the large coil (radius $r_2$) produces a magnetic field that is nearly uniform over the tiny area of the small coil (radius $r_1$), because $r_1 \ll r_2$. That makes the flux calculation trivial — no integration needed.

Step-by-Step Solution

1. Set up the geometry and the plan.

We have two coaxial circular coils with the same centre. Let the large coil (radius $r_2$) carry a current $I_2$. The small coil (radius $r_1$) is so tiny that the magnetic field from the large coil is essentially the same at every point inside the small coil. Mutual inductance $M$ is defined by:

where $\Phi_1$ is the magnetic flux through the small coil due to $I_2$.

2. Find the magnetic field at the centre of the large coil.

For a single circular loop of radius $r_2$ carrying current $I_2$, the magnetic field at its centre is:

directed along the axis (by the right-hand rule). This is a standard result from the Biot–Savart law.

3. Why can we treat the field as uniform over the small coil?

Because $r_1 \ll r_2$, the small coil’s entire area lies very close to the centre of the large coil. The field of a circular loop varies slowly near the centre — the leading correction is of order $(r_1/r_2)^2$. Since $r_1/r_2$ is tiny, the field is constant to an excellent approximation. So: …

Method: Flux Linkage Approach

This method uses the definition of mutual inductance — the flux linked with one coil due to current in the other, per unit current.

Steps

Step 1: Identify the primary and secondary coils

• Let the larger coil (radius $r_2$) carry current $I_2$.
• The smaller coil (radius $r_1$) is placed coaxially inside it, with $r_1 \ll r_2$.

Step 2: Find the magnetic field at the centre due to the larger coil

For a circular coil of radius $r_2$, the field at its centre is:

This field is uniform over a very small area near the centre.

Step 3: Why can we treat the field as uniform over the small coil?

Since $r_1 \ll r_2$, the small coil lies in a region where the field from the large coil is approximately constant. This is the key approximation.

Step 4: Calculate the flux through the small coil

Flux through the small coil (area $\pi r_1^2$): …

Common Mistakes & How to Avoid Them

Mistake 1: Assuming both coils produce a non-uniform field at the other

The error: Students often try to calculate the magnetic field of the larger coil at the smaller coil's location using the full Biot-Savart law, or vice versa — leading to messy integrals.

Why it's wrong:

Since $r_1 \ll r_2$, the smaller coil is so tiny compared to the larger one that the magnetic field from the larger coil is approximately uniform over the area of the smaller coil.

How to avoid:

Always check the size condition first. If $r_1 \ll r_2$, treat the larger coil's field as constant over the smaller coil's area. This lets you use:

Then flux through the small coil is simply:

Mistake 2: Confusing which coil's flux to calculate

The error: Students compute flux through the large coil due to current in the small coil — which is much harder because the small coil's field is non-uniform over the large area.

Why it's wrong:

Mutual inductance is symmetric ($M_{12} = M_{21}$), but one direction is much easier to compute. Always choose the path where the field is uniform over the other coil's area.

How to avoid:

• Always pass current through the larger coil.
• Then calculate flux through the smaller coil.
• This gives $M$ directly, and you never need to compute the harder direction.

Mistake 3: Forgetting that mutual inductance is purely geometric

The error: Students leave $I$ (current) in the final expression for $M$.

Why it's wrong:

Mutual inductance depends only on geometry (radii, number of turns, relative positions) and the medium. Current cancels out:

How to avoid:

After writing $\phi = B \times A$, substitute $B = \frac{\mu_0 I}{2 r_2}$ and $A = \pi r_1^2$. The $I$ cancels, giving: …