A light straight slider 1 m long can slide over two straight long parallel rails, vertically kept. when a uniform magnetic field of 0.6 T is applied perpendicular to the plane of rails, the slider is found moving with a constant velocity.
At this instant power dissipated in rails and slider due to their resistance is 1.96 W. If slider has a mass of 0.2kg, then the terminal velocity of the slider is ________.

Moderate

Solution

Let v be the terminal velocity of the slider then, emf developed =e=Blv
$Power dissipated will be P = i^{2} R = {(\frac{e}{R})}^{2} \cdot R = e i$
When slider achieve terminal velocity, net downward force on slider is zero.
Or magnetic force = gravitational force
$\begin{array}{l} \Rightarrow B i l = m g \\ \Rightarrow i = \frac{m g}{B l} = \frac{0.2 \times 9.8}{1.0 \times 0.6} = 3.27 A \\ As power developed P = i e \\ \Rightarrow Emg induced, e = \frac{P}{i} \Rightarrow e = \frac{1.96}{3.27} = 0.6 V \\ But e = B l v \Rightarrow v = \frac{e}{B l} = \frac{0.6}{0.6 \times 1.0} = 1 {ms}^{- 1} \end{array}$

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