Question

Two coplanar and concentric circular loops \( L_1 \) and \( L_2 \) are placed coaxially with their centres coinciding. The radii of \( L_1 \) and \( L_2 \) are 1 cm and 100 cm respectively. Calculate the mutual inductance of the loops. (Take \( \pi^2 = 10 \))

Hint:

Note that \(\pi^2 = 10\) significantly simplifies calculations in electromagnetic problems involving circular dimensions, especially in theoretical physics or academic exercises where approximate values are acceptable for learning concepts.

Answer 1

Given:  Radius of \( L_1 \), \( r_1 = 1 \, cm = 0.01 \, m \)  Radius of \( L_2 \), \( r_2 = 100 \, cm = 1 \, m \)  \( \pi^2 = 10 \)  Permeability of free space, \( \mu_0 = 4\pi  10^{-7} \, H/m \) The mutual inductance \( M \) between two coaxial circular loops is given by: \[ M = \frac{\mu_0 \pi r_1^2 r_2^2}{2(r_1^2 + r_2^2)^{3/2}} \] Substitute the given values: \[ M = \frac{(4\pi \times 10^{-7}) \pi (0.01)^2 (1)^2}{2((0.01)^2 + (1)^2)^{3/2}} \] Simplify the expression: \[ M = \frac{4\pi^2 \times 10^{-7} \times 10^{-4}}{2(0.0001 + 1)^{3/2}} \] Calculate the denominator: \[ (0.0001 + 1)^{3/2} = (1.0001)^{3/2} \approx 1^{3/2} = 1 \] Substitute \( \pi^2 = 10 \): \[ M = \frac{4 \times 10 \times 10^{-7} \times 10^{-4}}{2 \times 1} = \frac{4 \times 10^{-10}}{2} = 2 \times 10^{-10} \, H} \] Therefore, the mutual inductance of the loops is: \[ \boxed{2 \times 10^{-10} \, H}} \]