The rectangualr wire- frame, shown in has a width d, mass m, resistance R and a large length. A uniform magnetic field B exists to the left of the frame. A constant force F starts pushing the frame into the magnetic field at t =0. (a) Find the acceleration of the frame when its speed has increased to v. (b) Show that after some time the frame will move with a constant velocity till the whole frame enters into the magnetic field. find this velocity `v_0`. (c) show that the velocity at tiem t is given by ` v= v_0(1 - e^(-(Ft)/(mv_0)))`.
(##HCV_VOL2_C38_E01_083_Q01##)

Answer

(a)`(F-(B^(2)d^(2)v)/(R ))/(m)`; (b) `(FR)/(B^(2)d^(2))`; (c) `(RF)/(B^(2)d^(2))(1 - e^(-(B^(2)d^(2)t)/(mR)))`

Explanation

(a) `F_(net) = F - IdB`, `I = (e)/( R ) = (Bvd)/(R )`

`m xx a = F - (BdvdB)/(R )`, `a = ((F -B^(2)d^(2)v)/(R ))/(m)`
(b) at the stage of terminal velocity, `a = 0`, `v = v_(T)`
`(B^(2)d^(2)v_(T))/(R ) = F`, `v_(T) = (FR)/(B^(2)d^(2))`
( c) `(dv)/(dt) = ((F - B^(2)d^(2)v)/(R ))/(m)` `rarr` `int_(0)^(v) (dv)/(F - (B^(2)d^(2)v)/(R )) = int_(0)^(t)(dt)/(m)`
`1n[(F - (B^(2)d^(2)v)/(R ))/(F)] = (-B^(2)d^(2)t)/(mR)`
`1 - (B^(2)d^(2)v)/(RF) = e^((-B^(2)d^(2)t)/(mR ))`
`v = (RF)/(B^(2)d^(2))(1 - e^((-B^(2)d^(2)t)/(mR )))`