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).

Answer 1

Given:
Width of rectangular frame = d
Mass of rectangular frame = m
Resistance of the coil = R

(a) As the frame attains the speed v

Emf developed in side AB = Bdv (When it attains a speed v)

Current = $\frac{\mathrm{B}dv}{\mathrm{R}}$
The magnitude of the force on the current carrying conductor moving with speed v in direction perpendicular to the magnetic field as well as to its length is given by
$F=ilB$
Therefore, Force F_B = $\frac{\mathrm{B}{d}^{2}v}{\mathrm{R}}$

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

Therefore,Net force is given by
$$\begin{gathered} F_{\mathrm{net}}\mathit{}\mathit{=}\mathit{}F\mathit{-}{F}_B \\ {F}_{\mathrm{net}}\mathit{=}F-\frac{\mathrm{B}{d}^2{v}^2}{\mathrm{R}}=\frac{\mathrm{RF}-\mathrm{B}{d}^{2}v}{\mathrm{R}} \end{gathered}$$
Applying Newton's second law
$\frac{\mathrm{RF}-\mathrm{B}{d}^2{v}^2}{\mathrm{R}}=ma$
Net acceleration is given by a= $\frac{RF-B{d}^{2}v}{mR}$
(b)

Velocity of the frame becomes constant when its acceleration becomes 0.

Let the velocity of the frame be v₀
$$\begin{gathered} \frac{\mathit{F}}{\mathit{m}}\mathit{-}\frac{{\mathit{B}}^{\mathit{2}}{\mathit{d}}^{\mathit{2}}{\mathit{v}}_{\mathit{0}}}{\mathit{m}\mathit{R}}\mathit{=}\mathit{}\mathit{0} \\ \mathit{\Rightarrow }\frac{\mathit{F}}{\mathit{m}}\mathit{=}\frac{{\mathit{B}}^{{}_{\mathit{2}}}{\mathit{d}}^{\mathit{2}}{\mathit{v}}_{\mathit{0}}}{\mathit{m}\mathit{R}} \\ \mathit{\Rightarrow }{v}_{\mathit{0}}\mathit{=}\frac{\mathit{F}\mathit{R}}{{\mathit{B}}^{\mathit{2}}{\mathit{d}}^{\mathit{2}}} \end{gathered}$$
As the speed thus calculated depends on F, R, B and d all of them are constant, therefore the velocity is also constant.
Hence, proved that the frame moves with a constant velocity till the whole frame enters.
(c)
Let the velocity at time t be v.
The acceleration is given by
$$\begin{gathered} a=\frac{\mathrm{d}v}{\mathrm{d}t} \\ \Rightarrow \frac{RF-{\mathrm{d}}^2{\mathrm{B}}^2{v}^2}{m\mathrm{R}}=\frac{\mathrm{d}v}{\mathrm{d}t} \\ \frac{\mathrm{d}v}{\mathrm{RF}-d^2{\mathrm{B}}^2{v}^2}=\frac{\mathrm{d}t}{m\mathrm{R}} \\ \mathrm{Integrating} \\ \Rightarrow \underset{0}{\overset{v}{\int }}\frac{\mathrm{d}v}{\mathrm{RF}-d^2{\mathrm{B}}^2{v}^2}=\underset{0}{\overset{t}{\int }}\frac{\mathrm{d}t}{m\mathrm{R}} \\ \Rightarrow {\left[\ln (RF-d^2{\mathrm{B}}^{2}v)\right]}_0^v=-d^2{B}^2{\left[\frac{t}{Rm}\right]}_0^t \\ \Rightarrow \ln (RF-d^2{\mathrm{B}}^{2}v)-\ln \left(RF\right)=-\frac{d^2{B}^{2}t}{Rm} \\ \Rightarrow \frac{d^2{\mathrm{B}}^{2}v}{\mathrm{RF}}=1-e^{-\frac{d^2{\mathrm{B}}^{2}t}{\mathrm{R}m}} \\ v=\frac{FR}{l^2{\mathrm{B}}^2}\left(1-e^{-\frac{{\mathrm{B}}^2{d}^2{v}_{0}t}{\mathrm{R}{v}_{0}m}}\right) \\ v=v_0(1-e^{-Ft/v_{0}m})\left[\because F=\frac{B^2{d}^2{v}_0}{R}\right] \end{gathered}$$