Question

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 horiaontal, Fig. The circuit is closed through a perfert 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

The set up is shown in Fig. Component of weight of rod down the incline = mg sin `theta`. Component of magnetic field up the incline = B cos `theta`. Therefore, force on the rod due to magnetic field
`= B I l = (B cos theta) ((B cos theta d. upsilon))/(R ) d = (B^(2) d^(2) (cos theta)^(2) upsilon)/(R )`
`:.` The equation of motion of the rod can be written as
`m (d upsilon)/(dt) = m (d^(2) x)/(dt^(2)) = mg sin theta - (B^(2) d^(2))/(R ) (cos theta)^(2) upsilon`
`:. (d upsilon)/(dt) = g sin theta - (B^(2) d^(2))/(m R) (cos theta)^(2) upsilon`
or `(d upsilon)/(dt) + (B^(2) d^(2))/( mR) (cos theta)^(2) upsilon = g sin theta`
`:. upsilon = (g sin theta)/(B^(2) d^(2) cos^(2) theta//m r) + "A exp" [-(b^(2) d^(2))/(m R) (cos^(2) theta) t]`
where A is a constant that can be determined from initial conditions (A = 1).
`upsilon = (m R g sin theta)/(B^(2) d^(2) cos^(2) theta) [1 - exp (-(B^(2) d^(2))/(m R) (cos^(2) theta) t)]`