JEE Advanced 2019 · Paper 1 · Physics Q6

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PhysicsMultiple correct+4 / -1Question 6 of 54

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 $\vec B = B_0\left(1 + \left(\frac yL\right)^\beta\right)\hat k$ , 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:

The wire lies along y = x² from the origin to (√L, L) in the x-y plane; it moves with velocity V = V₀î in a magnetic field B pointing out of the page (along +z).

|\Delta\phi| = \frac12B_0V_0L

for

\beta = 0

|\Delta\phi| = \frac43B_0V_0L

for

\beta = 2

|\Delta\phi|

remains the same if the parabolic wire is replaced by a straight wire,

y = x

initially, of length

\sqrt2L

|\Delta\phi|

is proportional to the length of the wire projected on the

y

-axis.

Answer and worked solution

Official answer: B, C, D

Motional emf. The force per unit charge is $\vec v\times\vec B=V_0B(y)\,(\hat i\times\hat k)=-V_0B(y)\hat j$ , so only the $y$ -extent of the wire matters:
$|\Delta\phi|=V_0\int_0^LB_0\left(1+\left(\frac yL\right)^\beta\right)dy=B_0V_0L\left(1+\frac1{\beta+1}\right).$

(A) For $\beta=0$ this is $2B_0V_0L$ , not $\frac12B_0V_0L$ . False.
(B) For $\beta=2$ it is $\frac43B_0V_0L$ . True.
(C) The straight wire $y=x$ of length $\sqrt2L$ also spans $y$ from 0 to $L$ , so the emf is the same. True.
(D) For a given $\beta$ the emf depends only on the projection on the $y$ -axis. True.

(B), (C), (D)

Topics:Motional EMF Force on a moving charge

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