Two circular loops, one of small radius r and other of larger radius R, such thatR >> r, are placed coaxially with centres coinciding. Obtain the mutual inductance ofthe arrangement.

The mutual inductance of two coaxial loops can be calculated using the formula:

M = μ₀ * N₁ * N₂ * A₂ / l

where:

• M is the mutual inductance
• μ₀ is the permeability of free space (4π x 10^-7 T m/A)
• N₁ and N₂ are the number of turns in the first and second loop respectively
• A₂ is the area of the second loop
• l is the length of the solenoid

In this case, since we are dealing with single-turn loops (N₁ = N₂ = 1), the formula simplifies to:

M = μ₀ * A₂ / l

The area A₂ of the second loop (the larger one) can be calculated using the formula for the area of a circle (πR²), and the length l is the distance between the loops. Since the loops are coaxial and their centres coincide, we can consider l to be approximately equal to R (the radius of the larger loop).

So, the mutual inductance becomes:

M = μ₀ * πR² / R = μ₀ * πR

However, since R >> r, the magnetic field due to the smaller loop at the location of the larger loop can be considered uniform. Therefore, the mutual inductance can also be calculated as the product of the magnetic field due to the smaller loop and the area of the larger loop. The magnetic field B due to a current I in a single-turn loop of radius r at a distance x from the centre of the loop along its axis is given by:

B = μ₀ * I * r² / 2 * (r² + x²)^(3/2)

At x = R, this becomes:

B = μ₀ * I * r² / 2 * (r² + R²)^(3/2)

Since R >> r, we can neglect r² in comparison to R², so this simplifies to:

B = μ₀ * I * r² / 2R³

The mutual inductance is then given by:

M = B * A₂ = μ₀ * I * r² * πR² / 2R³ = μ₀ * I * πr² / 2R

This is the mutual inductance of the arrangement of the two loops.