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 overrightarrow 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:

Δ φ is proportional to the length of the wire projected on the y -axis.

Δ φ remains the same if the parabolic wire is replaced by a straight wire, y = x initially, of length √(2) L

Physics · Electromagnetic Induction

JEE Advanced 2019 — Paper 1 — Question 5

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 $\overrightarrow{\mathrm{B}}=\mathrm{B}_{0}\left(1+\left(\frac{\mathrm{y}}{\mathrm{L}}\right)^{\beta}\right) \hat{\mathrm{k}}$ , as shown in figure. If $\mathrm{V}_{0}$ , $\mathrm{B}_{0}, \mathrm{~L}$ 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:

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

Answer: A, B, D

Step-by-step solution

These is no change in flux through the loop OABO due to the movement of loop.

So potential difference developed in curved wire and the straight wire OA is same.

For $\beta=0,|\Delta \phi|=2 \mathrm{~B}_{0} \mathrm{~V}_{0} \mathrm{~L}$

For $\beta=2,|\Delta \phi|=\int_{0}^{L} B_{0}\left(1+\frac{y^{2}}{L^{2}}\right) V_{0} d y$

$=\frac{4}{3} \mathrm{~B}_{0} \mathrm{~V}_{0} \mathrm{~L}$

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2019
Paper: Paper 1
Subject: Physics
Chapter: Electromagnetic Induction
Topic: Motional EMF