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Question

Two ends of an inductor of inductance $$L$$ are connected to two parallel conducting wires. A rod of length $$l$$ and mass $$m$$ is given velocity $${v}_{0}$$ as shown. The whole system is placed in perpendicular magnetic field $$B$$. Find the maximum current in the inductor.
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A
mv0L
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B
mLv0
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C
mv02L
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D
None of these
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Solution

The correct option is B $$\sqrt { \cfrac { m }{ L } } { v }_{ 0 }$$
In the given question, as the magnetic filed is perpendicular to the velocity of the rod and the length of the rod is perpendicular to both the magnetic field and the velocity, $$ E=Blv $$
At any time $$ t $$, let the current in the circuit be $$ i $$ and the velocity of the wire be $$ v $$.

By Kirchoff's Loop law, we can write $$ Blv=-L\dfrac{di}{dt} $$

When current is maximum, $$ \dfrac{di}{dt}=0 \Rightarrow v=0 $$

So, the kinetic energy of the rod is completely converted into electromagnetic energy of the inductor.

Therefore, if the maximum current is $$ i_0 $$, we can write $$ \dfrac 12 Li_0^2=\dfrac 12mv_0^2 $$

So, $$ i_0=\sqrt{\dfrac mL}v_0 $$

Physics
NCERT
Standard XII

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