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 $\vec{B} = B_{0} (1 + {(\frac{y}{L})}^{β}) \hat{k},$ 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:

Question

Aicharya · Accepted Answer

For calculating the motional emf across the length of the wire, let us project wire such that

\vec{B}, \vec{v} -, \vec{i}

becomes mutually orthogonal. Thus, small emf 'dc' induced in elemental length

‘ d y ’

of projected wire, will be -

\Rightarrow d ε = B v_{0} d y = B_{0} [1 + {(\frac{y}{L})}^{β}] V_{0} d y

ε = \int_{0}^{L} B_{0} (1 + {(\frac{y}{L})}^{β}) V_{0} d y = \int_{0}^{L} B_{0} \cdot V_{0} \cdot d y + \int_{0}^{L} \frac{B_{0} \cdot V_{0}}{L^{β}} y^{β} \cdot d y

= B_{0} V_{0} L [1 + \frac{1}{β + 1}]

∴

emf in loop is proportional to

L

for given value of

β

.

\Rightarrow

Option

(A)

is correct.
for

β = 0; ε = 2 B_{0} V_{0} L

β = 2; ε = B_{0} V_{0} L [1 + \frac{1}{3}] = \frac{4}{3} B_{0} V_{0} L \Rightarrow

Option

(B)

is correct.
Now, the length of the projection of the wire

y = x

and length

\sqrt{2} L

on the y-axis is

L

thus the answer remain unchanged

\Rightarrow

Option

(C)

is correct.

A conducting wire of parabolic shape, initially y = x 2 , is moving with velocity V → = V 0 i ^ in a…

Solution