# Whats the resistance of inductor ?

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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. (neglect gravity and friction)

What I have tried is that potential difference created by rod would be $Blv_0$ so maximum current can be after inductor fully charged and that would be $Blv_0/R$ where R is resistance of the inductor. But that's not given in the question and rather the answer given is $\sqrt {m/L}×v_0$.

edited Mar 13

Initially I thought the rod was being pushed continuously so that it moves with constant velocity. But then the question does not make sense, as you found out.

Instead I think the rod is set in motion by giving it an impulse $mv_0$. When the velocity of the rod is $v$ there is an induced emf $V=Blv$ which causes a current $I$ to flow round the loop. While the rod is moving there is a force $BIl$ opposing its motion, slowing it down.

The value of $I$ changes during the rod's motion. While the rod is slowing down, the current is building up. When the rod has stopped moving the current gets it moving again in the opposite direction. As the speed of the rod increases the current decreases.

The question gives no value for resistance, so I think we must assume it is zero. We are also told to ignore friction. Therefore energy is conserved : the kinetic energy of the rod is transformed into magnetic energy stored in the inductor. There is an oscillation of energy between the inductor and the rod. The rod oscillates along the rail, forward and back.

$\frac12 mv_0^2=\frac12 mv^2+\frac12 LI^2$

The maximum current in the circuit occurs when the rod is instantaneously at rest, $v=0$.

Practically there will be some friction and some resistance, however small. So the oscillations will eventually die out. This is a linear system, equivalent to a damped harmonic oscillator.

answered Mar 10, 2017 by (28,746 points)
selected Mar 12, 2017
ohh I ignored thinking the constant force point and I got tricked with the question
:) :)
My initial reaction to the question was the same as yours. I thought it must be a mistake. Then I looked at the answer $I=v_0\sqrt{\frac{m}{L}}$ and wondered how that could be obtained.