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A rod of mass m and resistance R slides smoothly over two parallel perfectly conducting wires kept sloping at an angle \theta with respect to the horizontal (figure). The circuit is closed through a perfect conductor at the top. There is a constant magnetic field B along the vertical direction. If the rod is initially at rest, find the velocity of the rod as a function of time.

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The component of magnetic field along the inclined plane will be =B\sin \theta and other will be perpendicular i.e. =B\cos \theta. The conductor is moving perpendicular to =B\cos \theta. It is the vertical component of the magnetic field. The movement will cause motional emf across the two ends of the rod.

given by =v(B \cos \theta )d

This makes flow of induced current

i=\frac{v(B \cos \theta )d}{R}

where R is the resistance of rod. Now, current-carrying rod experience a magnetic force which is given by

F_{m}=iBd (horizontally in backward direction).

Now, the component of magnetic force parallels to the inclined plane along the upward direction.

F_{\parallel }=F_{m} \cos \theta =iBd \cos \theta =\left ( \frac{v(B \cos \theta )d}{R} \right )Bd \cos \theta



Also, the component of weight (mg) parallel to the inclined plane along downward direction = mg\; \sin \theta .

Now, by Newton's second law of motion

m\frac{d^{2}x}{dt^{2}}=mg \sin \theta -\frac{B \cos \theta d}{R}\left ( \frac{dx}{dt} \right )\times (BD)\; \cos \theta

\Rightarrow \frac{dv}{dt}=g\; \sin \theta -\frac{B^{2}d^{2}}{mR}(\cos \theta )^{2}v

\Rightarrow \frac{dv}{dt}+\frac{B^{2}d^{2}}{mR}(\cos \theta )^{2}v=g\; \sin \theta

But, this is the linear differential equation.

On solving, we get

v=\frac{g \sin \theta }{\frac{B^{2}d^{2}\cos^{2}\theta }{mR}}+ A\; exp\left ( -\frac{B^{2}d^{2}}{mR}(\cos ^{2}\theta )t \right )

A is a constant to be determined by initial conditions.

The required expression of velocity as a function of time is given by

\frac{mgR \sin \theta }{B^{2}d^{2}\cos^{2}\theta }\left ( 1-exp\left ( -\frac{B^{2}d^{2}}{mR}(\cos^{2}\theta )t \right ) \right )



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