99 Percentile Qs Bank for JEE MainPhysicsElectromagnetic Induction
Consider a toroid with rectangular cross section, of inner radius a , outer radius b and height h , carrying n number of turns. Then the self-inductance of the toroidal coil when current I passing through the toroid is
Options
- A₀ n^2 h 2 ( b a )
- B₀ n h 2 ( b a )
- C₀ n^2 h 2 ( a b )
- D₀ n h 2 ( a b )
Correct answer
A. ₀ n^2 h 2 ( b a )
Step-by-step solution
Given, a toroid with a rectangular cross-section of inner radius a and outer radius b . Height of the solenoid =h Magnetic field inside a rectangular toroid is given by B= ₀ n I 2 r using the infinitesimal cross-sectional area element, d x=h d r Flux passing through the cross-section of toroid. aligned & = B d x= _a^b ₀ n I 2 r (h d r) & = ₀ n I h 2 _a^b 1 r d r & = ₀ n h I 2 [ r l_a^b . & = ₀ n h I 2 ( b- a) & = ₀ n h I 2 ( b a ) aligned Now, self inductance of rectangular toroid, L= n I Putting the value of , we