JEE Advanced2019PhysicsElectromagnetic InductionActual
A conducting wire of parabolic shape, initially y = x 2 , is moving with velocity V → = V 0 i ^ in a non-uniform magnetic field B → = B 0 1 + y L β k ^ , as shown in figure. If V 0 , B 0 , L and β are positive constants and Δ ϕ is the potential difference developed between the ends of the wire, then the correct statement(s) is/are:
Options
- AΔ ϕ is proportional to the length of the wire projected on the y-axis.
- BΔ ϕ = 4 3 B 0 V 0 L for β = 2
- CΔ ϕ remains the same if the parabolic wire is replaced by a straight wire, y = x initially, of length 2 L
- DΔ ϕ = 1 2 B 0 V 0 L for β = 0
Correct answer
A. Δ ϕ is proportional to the length of the wire projected on the y-axis.
Step-by-step solution
For calculating the motional emf across the length of the wire, let us project wire such that B → , v → - , i → becomes mutually orthogonal. Thus, small emf 'dc' induced in elemental length ‘ d y ’ of projected wire, will be - ⇒ d ε = B v 0 d y = B 0 1 + y L β V 0 d y ε = ∫ 0 L B 0 1 + y L β V 0 d y = ∫ 0 L B 0 ⋅ V 0 ⋅ d y + ∫ 0 L B 0 ⋅ V 0 L β y β ⋅ d y = B 0 V 0 L 1 + 1 β + 1 ∴ emf in loop is proportional to L for given value of β . ⇒ Option A is correct. for β = 0 ; ε = 2 B 0 V 0 L β = 2 ; ε = B 0 V 0 L 1 + 1 3