JEE Advanced · Physics · 21. EMI
A long perfectly conducting wire is moving, with a velocity on a pair of horizontal rails of zero resistance. One side of the rails is connected to an inductor and a resistance as shown in figure. The horizontal rails, and lie in the same plane with a uniform magnetic field perpendicular to the plane. If the key is closed at certain instant, the current in the circuit after milli second is where the value of is_______.
[Assume the velocity of wire remains constant after key is closed. Given: where is base of the natural logarithm]

- A 0.32
- B 0.63
- C 0.45
- D 0.75
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 \(L - R \text { circuit } i=\frac{\varepsilon}{k}\left(1-e^{-t \cdot \frac{R}{L}}\right)] \)
\( i=10^{-3}(1-0.37) \)
\( i=0.63 mA\)
\(\varepsilon=10^{-3} \text { volt } \)
\( \Rightarrow i=\frac{10^{-3}}{1}\left(1-e^{-1}\right)\)
\(\Rightarrow\) \([\)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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