Question

A conducting wire of parabolic shape, initially $y = x^2$, is moving with velocity $\vec{V}=V_{0}\hat{i}$ a non-uniform magnetic field $\vec{B}=B_{0}\left(1+\left(\frac{y}{L}\right)^{\beta}\right)\hat{k},$ as shown in figure. If $V_0, B_0, L$ and $\beta$ are positive constants and $\Delta\phi$ is the potential difference developed between the ends of the wire, then the correct statement(s) is/are:

• A. $\left|\Delta \phi \right|=\frac{1}{2}B_{0}V_{0}L$ for $\beta=0$
• B. $\left|\Delta \phi \right|=\frac{4}{3}B_{0}V_{0}L for \beta=2$
• C. $\left|\Delta \phi \right|$ remains the same if the parabolic wire is replaced by a straight wire, $y = x$ initially, of length $\sqrt{2}L$
• D. $\left|\Delta \phi \right|$ is proportional to the length of the wire projected on the $y$ -axis

Answer 1

Answer: (D)

$\left|\Delta \phi \right|$ is proportional to the length of the wire projected on the $y$ -axis

Answer 2

$y=x^{2}, V=V_{0}\hat{i}, B=B_{0}\left(1+\left(\frac{y}{2}\right)^{\beta}\right)\hat{k}$
end points are (0, 0) and $\left(\sqrt{L}, L\right)$
Let at distance 'y' small length in y direction be dy
$\therefore d\varepsilon=V_{0}B\,dy$
$\therefore d\varepsilon=V_{0}B_{0}\left(1+\left(\frac{y}{L}\right)^{\beta}\right)dy=V_{0}B_{0}$ $\left[\int\limits^{{L}}_{{0}} y+\frac{y^{\beta+1}}{\left(\beta+1\right)L^{\beta}}]^{^{^L}}_{_{_{_0}}}\right]$
$\varepsilon=V_{0}B_{0}\left[L+\frac{L^{\beta+1}}{\left(\beta+1\right)L^{\beta}}\right] \Rightarrow \varepsilon=V_{0}B_{0}L\left(\frac{\beta+2}{\beta+1}\right)$
If $\beta=2$ then $\varepsilon=V_{0}B_{0}L$