Shown in the figure are two horizontal parallel conducting rails separated by distance $l$ . A uniform magnetic field B exists in vertical upward direction. A wire of mass $m$ can slide on the rails. The rails are connected to a source, which drives current $i$ in the circuit given by $i = a t - b t^{2}$ between $t = 0$ and $t = \frac{a}{b}$ , where $a$ and $b$ are positive constants. Coefficient of friction between the rails and the wire is $μ$ . Minimum $μ$ so that wire does not slide is $\frac{B a^{2} l}{x b m g}$ . Then the value of $x$ is

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answer is 4.

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Detailed Solution

For wire not to slide on the rails,
   $f = F_{m a g n e t i c}$
Or $μ m g = B i l$
$∴ μ \frac{B i l}{m g}$
   $= \frac{B (a t - b t^{2}) l}{m g}$
For minimum value of $μ, \frac{d μ}{d t} = 0$
Or    $a - 2 b t = 0$
Or    $t = \frac{a}{2 b}$
Or    $μ_{\min} = \frac{B a^{2} l}{4 b m g}$