Two long, horizontal pair of rails shown in the figure is connected using resistance R. The distance between the rails is l, the electrical resistance of the rails is negligible.
A conducting wire of mass m and length l can slide without friction on the pair of rails, in a vertical, homogeneous magnetic field of induction B.
A force of magnitude F₀ is exerted for sufficiently long time onto the conducting wire, so that the speed of the wire becomes nearly constant. The force F₀ is now removed
at a certain point P. What distance (in x 10² m) does the conducting wire cover on rails from point P before stopping?
$(Given: F_{0} = 20 N, m = 1.0 gm, R = 0.01 Ω, l = 10 cm, B=0.1T)$

Moderate

Solution

$For constant velocity, A = 0$
$\begin{array}{l} F_{0} = F_{m} = i l B = (\frac{ε}{R}) l B = (\frac{B l v_{0}}{R}) l B \\ v_{0} = \frac{F_{0} R}{B^{2} l^{2}} gives velocity at point ' P' \\ Now, retardation a = \frac{F_{m}}{m} = \frac{i l B}{m} \\ a = \frac{B^{2} l^{2}}{m R} v \Rightarrow - v \frac{d v}{d s} = \frac{B^{2} l^{2}}{m R} v \\ - \int_{v_{0}}^{0} d v = \frac{B^{2} l^{2}}{m R} \int_{0}^{s} d s \Rightarrow v_{0} = \frac{B^{2} l^{2}}{m R} s \\ s = \frac{m R v_{0}}{B^{2} l^{2}} = \frac{F_{0} m R^{2}}{B^{4} l^{4}} = 200 m \end{array}$

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