Two long parallel wires of zero resistance are connected to each other by a battery of 1.0 V as shown in figure. The separation between the wires is 0.5 m. A metallic bar, which is perpendicular to the wire and of resistance 10 Ω. moves on these wires when a magnetic field of 0. 02 tesla is acting perpendicular to the plane containing the bar and the wires. The velocity of the bar as a function of time (if the mass of the bar is 0.002 kg and at t = 0, v = 0 ) is:

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$V = 100 (1 - e^{- t / 200})$

$V = 100 (1 - e^{- 200 t})$

$V = 200 (1 - e^{- 500 t})$

$V = 100 (1 - e^{- t / 50})$

answer is A.

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Detailed Solution

$\begin{array}{l} i = \frac{ε - VBℓ}{R} \\ F = iBℓ = (\frac{ε - VBℓ}{R}) Bℓ \end{array}$

$\begin{array}{l} m \frac{d V}{dt} = \frac{(ε - VBℓ) Bℓ}{R} \\ \int_{0}^{V} \frac{d V}{ε - VBℓ} = \frac{Bℓ}{mR} \int_{t = 0}^{t} dt \\ - \frac{1}{Bℓ} \ln (\frac{e - VBℓ}{ε}) = \frac{Bℓ}{mR} t \\ V = \frac{ε}{Bℓ} (1 - e^{- \frac{B^{2} t^{2}}{mR} t}) \\ V = 100 (1 - e^{- t / 200}) \end{array}$