MHT CET Medical202622 April 2026Evening ShiftPhysicsElectromagnetic InductionActual
The self-inductance L of a toroid having circular cross-section of radius r and major radius R is ( r R )
Options
- A₀ N^2 R^2 r^2
- B₀ N^2 R^2 2r
- C₀ N^2 r^2 R^2
- D₀ N^2 r^2 2R
Correct answer
D. ₀ N^2 r^2 2R
Step-by-step solution
The magnetic field inside a toroid of major radius R is given by B = ₀ N I 2 R , assuming r R so that the field is approximately uniform across the cross-section. The cross-sectional area of the toroid is A = r^2 . The magnetic flux through a single turn is = B A = ( ₀ N I 2 R ) ( r^2) = ₀ N I r^2 2 R . The total flux linkage for N turns is = N = ₀ N^2 I r^2 2 R . The self-inductance is L = I = ₀ N^2 r^2 2 R . Answer: ₀ N^2 r^2 2R