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

Question

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:

TestHub · Accepted Answer

For calculating the motional emf across the length of the wire, let us project wire such that $\vec{B}, \vec{v} -, \vec{i}$ becomes mutually orthogonal. Thus, small emf 'dc' induced in elemental length $‘ d y ’$ of projected wire, will be -
$\Rightarrow d ε = B v_{0} d y = B_{0} [1 + {(\frac{y}{L})}^{β}] V_{0} d y$
$ε = \int_{0}^{L} B_{0} (1 + {(\frac{y}{L})}^{β}) V_{0} d y = \int_{0}^{L} B_{0} \cdot V_{0} \cdot d y + \int_{0}^{L} \frac{B_{0} \cdot V_{0}}{L^{β}} y^{β} \cdot d y$
$= B_{0} V_{0} L [1 + \frac{1}{β + 1}]$
$∴$ emf in loop is proportional to $L$ for given value of $β$ .
$\Rightarrow$ Option $(A)$ is correct.
for
$β = 0; ε = 2 B_{0} V_{0} L$
$β = 2; ε = B_{0} V_{0} L [1 + \frac{1}{3}] = \frac{4}{3} B_{0} V_{0} L \Rightarrow$ Option $(B)$ is correct.
Now, the length of the projection of the wire $y = x$ and length $\sqrt{2} L$ on the y-axis is $L$ thus the answer remain unchanged $\Rightarrow$ Option $(C)$ is correct.

Physics - EMI/AC Question with Solution | TestHub

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