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

Δ ϕ = ∫ V 0 B 0 1 + y ℓ β ⋅ d y ⇒ Δ ϕ = B 0 V 0 y + ( y ) β + 1 ( β + 1 ) ℓ β 0 ℓ ⇒ Δ ϕ = B 0 V 0 ℓ + ℓ β + 1 ⇒ Δ ϕ = B 0 V 0 β + 2 β + 1 ℓ If , β = 0 ⇒ Δ ϕ = 2 β 0 V 0 ℓ If , β = 2 ⇒ Δ ϕ = B 0 V 0 4 3 ℓ The length of the projection of the wire y = x of length 2 L on the y-axis is L thus the answer remain unchanged.

Δ ϕ = 1 2 B 0 V 0 L f o r β = 0

Δ ϕ = 4 3 B 0 V 0 L f o r β = 2

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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

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

The length of the projection of the wire y = x of length $\sqrt{2} L$ on the y-axis is $L$ thus the answer remain unchanged.

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