Question

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

$\overrightarrow{B}=B_0\left(1+{\left(\frac{y}{L}\right)}^{\beta }\right)\hat{k}$

as shown in figure. If $V_0$, $B_0$, $L$and $B$ are $+ve$constants and$∆\varphi$ is potential difference developed between the ends of the wire, then correct statement(s) is/are

• A. $\left|\Delta \phi \right|=(1/2)B_0{V}_{0}Lfor\beta =0$
• B. $\left|\Delta \phi \right|=(4/3)B_0{V}_{0}Lfor\beta =2$
• C. $\left|\Delta \phi \right|$ is proportional to the length of wire projected on y-axis.
• D. $\left|\Delta \phi \right|$ remains same if the parabolic wire is replaced by a straight wire, $y=x$, initially of length $\sqrt{2}$

Answer 1

Answer: (B), (C), (D)

$\left|\Delta \phi \right|$ remains same if the parabolic wire is replaced by a straight wire, $y=x$, initially of length $\sqrt{2}$

Step 1 : Given data

Wire's shape $y=x^2$

Velocity $\overrightarrow{V}=V_0\hat{i}$

Magnetic field $\overrightarrow{B}=B_0\left(1+{\left(\frac{y}{L}\right)}^{\beta }\right)\hat{k}$

Potential differences developed between the ends $=∆\varphi$

$\epsilon =$emf of the wire

$L=$ length of the wire

Step 2: To find the correct statement for $∆\varphi$

Calculating the motional emf across the length of the wire, $\overrightarrow{B},\overrightarrow{V}$are mutually orthogonal, thus

$$\begin{gathered} d\epsilon =B{V}_{0}dy \\ =B_0\left[1+{\left(\frac{y}{L}\right)}^{\beta }\right]V_{0}dy \\ \epsilon ={\int }_0^L{B}_0\left(1+\frac{1}{\beta +1}\right) \end{gathered}$$

emf in the loop is proportional to $L$for given value of $\beta$

for

$$\begin{gathered} \beta =0;\epsilon =2B_0{V}_{0}L \\ \beta =2;\epsilon =B_0{V}_{0}L\left[1+\frac{1}{3}\right] \\ =\frac{4}{3}{B}_0{V}_{0}L \end{gathered}$$

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

Hence the correct answer is Option B , C , D