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  • A pair of parallel horizontal conducting rails of negligible resistance shorted at one end is fixed on a table. A constant magnetic field B exsists perpendicular to the table. T

A pair of parallel horizontal conducting rails of negligible resistance shorted at one end is fixed on a table. A constant magnetic field B exsists perpendicular to the table. The distance between the rails is L. A massless rod having resistance R can slide on the rails
frictionlessly. The rod is connected to the mass m by a light string passing over a light frictionless pulley as shown.Calculate the terminal velocity of the rod.

Option: 1

\frac{\mathrm{mgR}}{\mathrm{B}^2 \mathrm{l}^2}


Option: 2

\frac{2 \mathrm{mgR}}{\mathrm{B}^2 \mathrm{l}^2}


Option: 3

\frac{\mathrm{mgR}}{2 \mathrm{~B}^2 \mathrm{l}^2}


Option: 4

\frac{\mathrm{mgR}}{3 \mathrm{~B}^2 \mathrm{l}^2}


Answers (1)

best_answer

Induced emf in the rod \mathrm {=Blv}
\therefore Current in the circuit   \mathrm {=\frac{B l v}{R}}

\therefore Magnetic force on the rod  \mathrm {=Bil}

\mathrm {F=\frac{B^2 I^2 v}{R}}

(The force in directed opposite in the velocity from Lenz’s law)
As the rod is massless, the net force on it must be zero.

\therefore The tension in the string   \mathrm {=\frac{B^2 I^2 v}{R}}

Now applying Newton’s second law on mass m,

\mathrm {m g-\frac{B^2 l^2 v}{R}=m \frac{d v}{d t}}

At terminal velocity \frac{\mathrm{dv}}{\mathrm{dt}}=0\\

\Rightarrow \mathrm{v}_{\mathrm{tr}}=\frac{\mathrm{mgR}}{\mathrm{B}^2 \mathrm{l}^2}

Posted by

jitender.kumar

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