A conducting wire of parabolic shape, initially y = x2, is moving with velocity vecV=V0 hati a non-uniform magnetic field vecB=B0(1+((y/L))β) hatk, 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:

Question

A conducting wire of parabolic shape, initially y = x2, is moving with velocity vecV=V0 hati a non-uniform magnetic field vecB=B0(1+((y/L))β) hatk, 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: (A) |Δ φ |=(1/2)B0V0L for β=0 (B) |Δ φ |=(4/3)B0V0L for β=2 (C) |Δ φ | remains the same if the parabolic wire is replaced by a straight wire, y = x initially, of length √2L (D) |Δ φ | is proportional to the length of the wire projected on the y -axis

Tardigrade Admin · Accepted Answer

y=x2, V=V0 hati, B=B0(1+((y/2))β) hatk end points are (0, 0) and (√L, L) Let at distance 'y' small length in y direction be dy ∴ dε=V0B dy ∴ dε=V0B0(1+((y/L))β)dy=V0B0[∫ limitsL0 y+(yβ+1/(β+1)Lβ)]L0] ε=V0B0[L+(Lβ+1/(β+1)Lβ)] ⇒ ε=V0B0L((β+2/β+1)) If β=2 then ε=V0B0L

$∣ Δ ϕ ∣ = \frac{1}{2} B_{0} V_{0} L$ for $β = 0$

$∣ Δ ϕ ∣ = \frac{4}{3} B_{0} V_{0} L f or β = 2$

$∣ Δ ϕ ∣$ remains the same if the parabolic wire is replaced by a straight wire, $y = x$ initially, of length $2 L$

$∣ Δ ϕ ∣$ is proportional to the length of the wire projected on the $y$ -axis

∣Δϕ∣=21​B0​V0​L for β=0

∣Δϕ∣=34​B0​V0​Lforβ=2

∣Δϕ∣ remains the same if the parabolic wire is replaced by a straight wire, y=x initially, of length 2​L

∣Δϕ∣ is proportional to the length of the wire projected on the y -axis

$∣ Δ ϕ ∣ = \frac{1}{2} B_{0} V_{0} L$ for $β = 0$

$∣ Δ ϕ ∣ = \frac{4}{3} B_{0} V_{0} L f or β = 2$

$∣ Δ ϕ ∣$ remains the same if the parabolic wire is replaced by a straight wire, $y = x$ initially, of length $2 L$

$∣ Δ ϕ ∣$ is proportional to the length of the wire projected on the $y$ -axis