Qstnid -Iv-48-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 - 38: Electromagnetic Induction

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NEET-XII-Physics

38: Electromagnetic Induction

with Solutions - page 10

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#48
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
Ans : 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
`` {F}_{\,\mathrm{\,net\,}}\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\,}}``
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
`` \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{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}}}``
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
`` 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[\,\mathrm{\,ln\,}(RF-{d}^{2}{\,\mathrm{\,B\,}}^{2}v)\right]}_{0}^{v}=-{d}^{2}{B}^{2}{\left[\frac{t}{Rm}\right]}_{0}^{t}``
`` \Rightarrow \,\mathrm{\,ln\,}(RF-{d}^{2}{\,\mathrm{\,B\,}}^{2}v)-\,\mathrm{\,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]``
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#48-a
Find the acceleration of the frame when its speed has increased to v.
Ans : 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
`` {F}_{\,\mathrm{\,net\,}}\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\,}}``
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}``

#48-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.
Ans : Velocity of the frame becomes constant when its acceleration becomes 0.
Let the velocity of the frame be v₀
`` \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{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}}}``
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.

#48-c
Show that the velocity at time t is given by
v = v₀(1 - e^-Ft^/mv₀).
Figure
Ans : Let the velocity at time t be v.
The acceleration is given by
`` 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[\,\mathrm{\,ln\,}(RF-{d}^{2}{\,\mathrm{\,B\,}}^{2}v)\right]}_{0}^{v}=-{d}^{2}{B}^{2}{\left[\frac{t}{Rm}\right]}_{0}^{t}``
`` \Rightarrow \,\mathrm{\,ln\,}(RF-{d}^{2}{\,\mathrm{\,B\,}}^{2}v)-\,\mathrm{\,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]``
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