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

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.

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

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.

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A conducting wire of parabolic shape, initially $y = x^{2}$ is moving with velocity $\vec{V} = V_{0} \hat{i}$ in a non – uniform magnetic field $\vec{B} = B_{0} (1 + {(\frac{y}{L})}^{β}) \hat{k}$ , 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

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$|Δ ϕ| = \frac{1}{2} B_{0} V_{0} L f o r β = 0$

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

$|Δ ϕ|$ remains the same if the parabola wire is replaced by a straight wire, $y = x$ initially, of length $\sqrt{2} L$

$|Δ ϕ| = \frac{4}{3} B_{0} V_{0} L f o r β = 2$

answer is B, C, D.

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Detailed Solution

$\begin{array}{l} Δ ϕ = \int V_{0} B_{0} (1 + {(\frac{y}{ℓ})}^{β}) \cdot d y \\ \Rightarrow Δ ϕ = B_{0} V_{0} {[y + \frac{(y)^{β + 1}}{(β + 1) ℓ^{β}}]}_{0}^{ℓ} \\ \Rightarrow Δ ϕ = B_{0} V_{0} (ℓ + \frac{ℓ}{β + 1}) \\ \Rightarrow Δ ϕ = B_{0} V_{0} (\frac{β + 2}{β + 1}) ℓ \\ If, β = 0 \\ \Rightarrow Δ ϕ = 2 β_{0} V_{0} ℓ \\ If, β = 2 \\ \Rightarrow Δ ϕ = B_{0} V_{0} (\frac{4}{3}) ℓ \end{array}$