Most Important Selected Qs for JEE AdvancedPhysicsElectromagnetic Induction
Paragraph: A thin superconducting (zero resistance) ring is held above a vertical long solenoid, as shown in the figure. The axis of symmetry of the ring is same to that of the solenoid. The cylindrically symmetric magnetic field around the ring can be described approximately in terms of the vertical and radial component of the magnetic field vector as B_z=B_d(1- z) and B_r=B₀ r , where B₀, and are positive constants
Options
- A1 ~B ₀ r₀^2 2 ~mL
- B1 2 ~B ₀ r ₀^2 2 ~mL
- C1 B₀ r₀^2 ~mL 3
- Dring will not perform SHM Total magnetic flux at any position = B _ z r ₀^2- LI Since, R=0 , so =B₀(1- z) r₀^2
Correct answer
A. 1 ~B ₀ r₀^2 2 ~mL
Step-by-step solution
Total magnetic flux at any position = B _ z r ₀^2- LI Since, R=0 , so =B₀(1- z) r₀^2-L I= constant From initial condition (z=0, I=0) , the value of constant is = B ₀ r ₀^2 Using the above equation the current in the ring I = 1 ~L ~B ₀ r ₀^2 z The lorentz force acting on the ring (which can only be vertical, because of the symmetry of the assembly) can be expressed as F _ z =- B _ r I ( z ) 2 r ₀=- 2 ~B ₀^2 ^2 r ₀^4 z ~L =- kz Equation of motion of the ring is ma _ z = F _ z - mg =- kz - mg Equilibrium position z ₀=