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JEE Advanced · Physics · 21. EMI

A conducting wire of parabolic shape, initially y=x2, is moving with velocity V=V0i^ in a non-uniform magnetic field B=B01+yLβk^, as shown in figure. If V0, B0, L and β are positive constants and Δϕ is the potential difference developed between the ends of the wire, then the correct statement(s) is/are:

  1. A Δϕ is proportional to the length of the wire projected on the y-axis.
  2. B Δϕ=43B0V0L for β=2
  3. C Δϕ remains the same if the parabolic wire is replaced by a straight wire, y=x initially, of length 2L
  4. D Δϕ=12B0V0L for β=0
Verified Solution

Answer & Solution

Correct Answer

(A) Δϕ is proportional to the length of the wire projected on the y-axis.

Step-by-step Solution

Detailed explanation


For calculating the motional emf across the length of the wire, let us project wire such that B,v-,i becomes mutually orthogonal. Thus, small emf 'dc' induced in elemental length dy of projected wire, will be -
\(\Rightarrow d \varepsilon=B v_0 d y=B_0\left[1+\left(\frac{y}{L}\right)^\beta\right] V_0 d y \)
\( \varepsilon=\int_0^L B_0\left(1+\left(\frac{y}{L}\right)^\beta\right) V_0 d y=\int_0^L B_0 \cdot\)\( V_0 \cdot d y+\int_0^L \frac{B_0 \cdot V_0}{L^\beta} y^\beta \cdot d y \)
\( =B_0 V_0 L\left[1+\frac{1}{\beta+1}\right]\)
emf in loop is proportional to L for given value of β .
Option A is correct.
for
β=0;ε=2B0V0L
β=2;ε=B0V0L1+13=43B0V0L Option (B) is correct.
Now, the length of the projection of the wire y=x and length 2L on the y-axis is L thus the answer remain unchanged Option (C) is correct.
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