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A bar of mass `m` and length `l` moves on two frictionless parallel rails in the presence of a uniform magnetic field directed into the plane of the paper. The bar is given and initial velocity `v_(i)` to the right and released. Find the velocity of bar, induced emf across the bar and the current in the circuit as a function of time |
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Answer» The induced current in the counter clockwise direction and the magnetic force on the bar is given by `F_(B) = -ilB`. The negative sign indicates that the force is towards the left and retards motion. `F = m a` `-i lB = m.(dv)/(dt)` Because the force depends on current and the current depends on the speed, the force is not constant and the acceleration of the bar is not constant. The induced current is given by `i = (B l v)/(R )`, `-ilB = m.(d v)/(d t)` `-((Blv)/(R ))lB = m.(dv)/(dt) rArr (dv)/(dt) = -(B^(2)l^(2))/(mR)dt` `underset(v_(1))overset(v)int(dv)/(v) = -(B^(2)l^(2))/(mR)underset(0)overset(t)intdt`, `ln ((v)/(v_(1))) = -(B^(2)l^(2))/(mR)t = (-t)/(T)` where `T = (mR)/(B^(2)l^(2)) rArr v = v_(i)e^((t)/(T))` The speed of the bar therefore decreases exponentially with time under the action of magnetic retarding force. `emf = iR = Blv_(i)e^((t)/(T))`, current: `i = (Blv)/(R ) = (B l)/(R )v_(1)e^((t)/(T))` |
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