A horizontal rod of mass 2kg is kept touching two vertical parallel rough rails, carrying current. There is a magnetic field B = 2T present vertically downward. The rails are connected to battery of 100V at t = 0. The resistance of the circuit is $5 Ω$ and starts to increase at constant rate $0.5 Ω /s$ . The coefficient of friction between the rails and rod is $μ = 3/4 (g = 10 {m/s}^{2})$ and separation between the rails in 1m). Then

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Acceleration of rod at t = 0 is

5 {m/s}^{2}

The friction force acting on the rod at t = 0 is 30 N

At t = 5 sec. rod is about to start

The friction force acting on the rod at t = 0 is 20 N

answer is A, D.

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Detailed Solution

t = 0 i - \frac{Σ}{R} = \frac{100}{5} = 20

$F = BIL = 40 N$ from FLR, this force is equal to N

$F = μ N = \frac{3}{4} \times 40 = 30 N$

$mg = 20N (f > mg)$

$∴$ By self adjusting property f = 20 N

If $f = 20 N \Rightarrow μ N = 20 \Rightarrow μ Bil = 20$

$i = \frac{40}{3} = \frac{100}{R^{|}} \Rightarrow R^{|} = 7.5$

$∴ 5 + (0.5) t = 7.5 \Rightarrow t = 5 s$

At t = 5 s the rod is ready to start

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