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JEE Advanced · Physics · 21. EMI

A 10 cm long perfectly conducting wire PQ is moving, with a velocity 1 cm/s on a pair of horizontal rails of zero resistance. One side of the rails is connected to an inductor L=1 mH and a resistance R=1Ω as shown in figure. The horizontal rails, L and R lie in the same plane with a uniform magnetic field B=1 T perpendicular to the plane. If the key S is closed at certain instant, the current in the circuit after 1 milli second is x×10-3 A, where the value of x is_______.
[Assume the velocity of wire PQ remains constant 1cm/s after key S is closed. Given: e-1=0.37, where e is base of the natural logarithm]

  1. A 0.32
  2. B 0.63
  3. C 0.45
  4. D 0.75
Verified Solution

Answer & Solution

Correct Answer

(B) 0.63

Step-by-step Solution

Detailed explanation

\(\Rightarrow \varepsilon=(\vec{v} \times \vec{B}) \vec{d} l=10^{-2} \times 1 \times 10^{-1}\)\(\left[\begin{array}{c}\text { Motional emf induced in the rod } \\ \sum \text{ind} =(\vec{V} \times \vec{B}) \vec{d} l \\ =\text { B.V.l }\end{array}\right]\)
\(\varepsilon=10^{-3} \text { volt } \)
\( \Rightarrow i=\frac{10^{-3}}{1}\left(1-e^{-1}\right)\)
\(\Rightarrow\) \([\)Equation of growth of current in \(L - R \text { circuit } i=\frac{\varepsilon}{k}\left(1-e^{-t \cdot \frac{R}{L}}\right)] \)
\( i=10^{-3}(1-0.37) \)
\( i=0.63 mA\)
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