Two planar concentric rings of metal wire having radii ' r₁ ' and ' r₂ ' (with r₁ > r₂ ) are placed in air. The current 'I' is flowing through the coil of larger radius. The mutual inductance between the coils is given by ( ₀ = permeability of free space)

Step-by-step solution

Magnetic field at the center of the larger ring of radius r₁ carrying current I is given by B = ₀ I 2 r₁ Since r₁ > r₂ , the magnetic field over the area of the smaller ring can be considered uniform and approximately equal to the field at the center. Magnetic flux linked with the smaller ring of radius r₂ is = B A₂ = ( ₀ I 2 r₁ ) ( r₂^2) By definition of mutual inductance, = M I . Equating the two expressions for flux, we get M I = ₀ r₂^2 2 r₁ I M = ₀ r₂^2 2 r₁ Answer: ₀ , r₂^2 2 , r₁