The conducting rod `ab` shown in figure makes contact with metal rails `ca` and `db`. The apparatus is in a uniform magnetic field `0.800 T`, perpendicular to the plane of the figure.

(a) Find the magnitude of the emf induced in the rod when it is moving toward the right with a speed `7.50 m//s`.
(b) In what direction does the current flow in the rod?
(c) If the resistance of the circuit abdc is `1.50 Omega` (assumed to be constant), find the force (magnitude and direction) required to keep the rod moving to the right with a constant speed of `7.50 m//s`. You can ignore friction.
(d) Compare the rate at which mechanical work is done by the force `(Fu)` with the rate at which thermal energy is developed in the circuit `(I^2R)`.

Answer

A, B

Explanation

a. `e=Bvl=0.8xx7.5xx0.5=3V`
b. Current will flow in anti clockwise direction, as magnetic field in `ox` direction passing through the closed loop is increasing. Therefore, induced current will produce magnetic field in `o.` direction.
c. `F=F_m=ilB=e/R lB= ((Bvl)/R)Bl=(B^2l^2)/Rv`
`=((0.8)^2(0.5)^2)/1.5xx7.5=0.8N`
d. `Fv=0.8xx7.5=6w`
`i^2R=((Bvl)/R)^2R=(B^2l^2)/R.v^2`
`=((0.8)^2(0.5)^2)/1.5xx(7.5)^2=6W`
So we can see that both rates are equal.