Question

A rod of mass m and resistance R slides smoothly over two parallel perfectly conducting wires kept sloping at an angle θ 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.

Answer 1

Let us first divide the magnetic field in the components one is along the inclined plane = B sin θ and other component of magnetic field is perpendicular the plane = B cos θ

Now, the conductor moves with speed v perpendicular to B cos θ, component of magnetic field. This causes motional emf across two ends of rod, which is given by = v(B cos θ)d

(A)

(B)

This makes flow of induced current `i = (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 backwards direction). Now, the component of magnetic force parallel to the inclined plane in upward direction

`F_(||) = F_m cos θ = -Bd cos θ = ((v(B cos θ)d)/R) Bd cos θ`

Where, `v = (dx)/(dt)`

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

Now, by Newton's second law of motion

`m (d^2x)/(dl^2) = mg sin θ - (B cos θ d)/R ((dx)/(dt)) xx (Bd) cos θ`

⇒ `(dv)/(dt) = g sin θ - (B^2d^2)/(mR) (cos theta)^2v`

⇒ `(dv)/(dt) + (B^2d^2)/(mR) (cos θ)^2v = g sin θ`

But, this is the linear differential equation.

On solving, we get

`v = ((g sin θ)/(B^2d^2 cos^2 θ))/(mR) + A exp(- (B^2d^2)/(mR) (cos^2 θ)t)`

A is a constant to be determined by initial conditions.

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

= `(mgR sin θ)/(B^2d^2 cos^2θ) (1 - exp (- (B^2d^2)/(mR) (cos^2θ)t))`