Question

The rectangular wire-frame, shown in 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) 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 v0. (c) Show that the velocity at time t is given by
v = v0(1 − e−Ft/mv0).
Figure

Answer 1

Answer:

where is the figure of mentioned in the above line please

Answer 2

(a) The acceleration of the frame when its speed has increased to v is $\mathrm{a}=\frac{R F-B d^{2} v}{m R}$

(b) The velocity $v_0$ is $\frac{F R}{B^{2} d^{2}}$

(c) The velocity at time t is given by  v = v0(1 − e−Ft/mv0)

Explanation:

Right-angled panel width = d

Rectangular frame mass = m

Coil's resistance = R

(a) As the frame reaches speed v

Emf evolved sideways AB = Bdv (When it reaches a speed)

$\text { Current }=\frac{B d v}{R}$

The magnitude of the force on the current carrying conductor traveling at velocity in a perpendicular direction to the magnetic field as well as its length is determined by

$F=i l B$

Therefore, $\text { Force } F_{B}=\frac{B d^{2} v}{R}$

Since the force is in the opposite direction to that of the frame motion.

Therefore, total force is generated by

$F_{n e t}=F-F_{B}$

$F_{n o t}=F-\frac{B d^{2} v^{2}}{R}=\frac{R F-B d^{2} v}{R}$

Applying Second Law of Newton

$\frac{R F-B d^{2} v}{R}=m a$

Total acceleration $\mathrm{a}=\frac{R F-B d^{2} v}{m R}$

(b) Frame velocity becomes constant when its acceleration is 0.

Let the Frame velocity be $v_0$

$\frac{F}{m}-\frac{B d^{2} v_{0}}{m R}=0$

$\frac{F}{m}=\frac{B d^{2} v_{0}}{m R}$

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

Since the speed thus measured depends on F, R, B and d they are all constant, the velocity is therefore also constant.

Proved therefore that the frame shifts with a constant velocity until the whole frame joins.

(c)  The velocity at time t is given by  v = v0(1 − e−Ft/mv0)

Let the speed t be v at the time.

we know that

$a=\frac{d v}{d t}$

$\frac{R F-d^{2} B^{2} v^{2}}{m R}=\frac{d v}{d t}$

$\frac{d v}{R F-d^{2} B^{2} v^{2}}=\frac{d t}{m R}$

On Integrating, we get,

$\int_{0}^{v} \frac{d v}{R F-d^{2} B^{2} v^{2}}=\int_{0}^{t} \frac{d t}{m R}$

$\left[\ln \left(R F-d^{2} B^{2} v\right)\right]_{0}^{v}=-d^{2} B^{2}\left[\frac{t}{R m}\right]_{0}^{t}$

$\ln \left(R F-d^{2} B^{2} v\right)-\ln (R F)=-d^{2} B^{2} \frac{t}{R m}$

$\frac{d^{2} B^{2} v}{R F}=1-e^{-d^{2} B^{2}} \frac{t}{R m}$

      $v=\frac{F R}{l^{2} B^{2}}\left(1-e^{-d^{2} B^{2} \frac{t v_{0}}{R m} v_{0}}\right)$

        $=v_{0}\left(1-e^{\frac{-F t}{v_{0} m}}\right) \quad \because F=\frac{d^{2} B^{2} v_{0}}{R}$