A conducting wire of parabolic shape, initially y=x2 is moving with velocity V→=V0i^ in a non – uniform magnetic field B→=B01+yLβk^, as shown in figure. If V0.B0,L and β are positive constants and Δϕ is the potential difference developed between the ends of the wire, then the correct statement(s) is/are
see full answer
Your Exam Success, Personally Taken Care Of
1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.
An Intiative by Sri Chaitanya
a
Δϕ=12B0V0L for β=0
b
Δϕis proportional to the length of the wire projected on the y-axis.
c
Δϕremains the same if the parabola wire is replaced by a straight wire, y=x initially, of length 2L
d
Δϕ=43B0V0L for β=2
answer is B.
(Unlock A.I Detailed Solution for FREE)
Best Courses for You
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
Detailed Solution
Δϕ=∫V0B01+yℓβ⋅dy⇒Δϕ=B0V0y+(y)β+1(β+1)ℓβ0ℓ⇒Δϕ=B0V0ℓ+ℓβ+1⇒Δϕ=B0V0β+2β+1ℓ If ,β=0⇒Δϕ=2β0V0ℓ If ,β=2⇒Δϕ=B0V043ℓThe length of the projection of the wire y = x of length 2L on the y-axis is L thus the answer remain unchanged.