Question

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

• A. $|\Delta \varphi |=\frac{4}{3}{B}_0{V}_{0}L$ for $\beta =2$
• B. $|\Delta \varphi |$ remains same if the parabolic wire is replaced by a straight wire, $y=x,$ initially, of length $\sqrt{2}l$
• C. $|\Delta \varphi |=\frac{1}{2}{B}_0{V}_{0}L$ for $\beta =0$
• D. $|\Delta \varphi |$ is proportional to the length of wire projected on $y$ -axis

Answer 1

Answer: (A), (B), (D)

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


For calculating the motional emf across the length of the wire, let us project wire such that $\overrightarrow{B},\overrightarrow{v},\overrightarrow{l}$ becomes mutually orthogonal. Thus

$d\epsilon =B{v}_{0}dy=B_0[1+{\left(\frac{y}{L}\right)}^{\beta }]V_{0}dy$
$\epsilon ={\int }_0^L{B}_0(1+{\left(\frac{y}{L}\right)}^{\beta })V_{0}dy$
$=B_o{V}_o[y+\frac{y^{\beta +1}}{L^{\beta }(\beta +1)}{]}_0^L$
$=B_o{V}_o[L+\frac{L^{\beta +1}}{L^{\beta }(\beta +1)}]$
$=B_0{V}_{0}L[1+\frac{1}{\beta +1}]$ emf in loop is proportional to L for given value of $\beta$.
For
$\beta =0;\epsilon =2B_0{V}_{0}L$
$\beta =2;\epsilon =B_0{V}_{0}L[1+\frac{1}{3}]=\frac{4}{3}{B}_0{V}_{0}L$
Since the projection of length of the wire on the y-axis is $L$, therefore the answer remain unchanged