Question

The rectangular wire frame, shown in the figure 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 how that after some time the frame will move with a constant velocity till the whole frame enters into the magnetic field. Find the velocity $v_0$
(c) Show that the velocity at time t is given by$v=v_0(1-e^{\frac{-Ft}{m{v}_0}})$

Answer 1

Given :

width of the rectangular frame= d

mass of the rectangular frame= m

the resistance of the coil= R

(a) As the frame attains the speed v

emf developed in the side AB = Bdv ( when it attains a speed v )

$current=\frac{Bdv}{R}$

the magnitude of the force on the current carrying conductor moving speed v in direction perpendicular field as well as to length is given by

$F=ilB$

therefore, Force$F_s=\frac{B{d}^{2}v}{R}$ . . . . . . . . . . .(1)

as the force is in the direction opposite to that of the motion to the frame

therefore the net force is given by:

$F_{net}=F-F_B$

$F_{net}=F-\frac{B{d}^{2}v}{R}=\frac{RF-B{d}^{2}v}{R}$

applying Newton's second law,

$\frac{RF-B{d}^2{v}^2}{R}=ma$

net acceleration is given by $a=\frac{RF-B{d}^{2}v}{mR}$

(b) the velocity of the frame becomes constant when its acceleration becomes 0.

Let the velocity of the frame be $v_0$

$\frac{F}{m}-\frac{B^2{d}^2{v}_0}{mR}=0$

$\Rightarrow \frac{F}{m}=\frac{B^2{d}^2{v}_0}{mR}$

$\Rightarrow v_0=\frac{FR}{B^2{d}^2}$

As the speed thus calculated depends on F, R, B and all of them are constant, thus the velocity is also constant.

hence, it proves that the frame moves with a constant velocity until the whole frame enters

(c) Let the velocity at time$t$ be${}^{\prime }{v}^{\prime }$.

The acceleration is given by

$a=\frac{dv}{dt}$

$\Rightarrow \frac{RF-d^2{B}^2{v}^2}{mR}=\frac{dt}{mR}$

Integrating

$\Rightarrow {\int }_0^v\frac{RF-d^2{B}^2{v}^2}{mR}={\int }_0^t\frac{dv}{mR}$

$\Rightarrow {\left[\ln \right(RF-d^2{B}^{2}v\left)\right]}_0^v=-d^2{B}^2{\left[\frac{t}{Rm}\right]}_0^t$

$\Rightarrow ln(RF-d^2{B}^{2}v)-ln\left(RF\right)=-d^2{B}^2\frac{t}{Rm}$

$\Rightarrow \frac{d^2{B}^{2}v}{RF}=1-e^{-frac{d}^2{B}^{2}tRm}$

$v=\frac{FR}{l^2{B}^2}(1-e^{-\frac{B^2{d}^{2}t}{R{v}_{0}m}})$

$v=v_0(1-e^{-\frac{FT}{v_{0}m}})$ from (1)