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. 1. Find the acceleration of the frame when its speed has increased to v. 2. 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. 3. Show that the velocity at time t is given by v=v_0 (1-e^ - ft mv_0 )

1. emf developed = Bdv (when it attains a speed v) Current = Bdv R Force = Bd^2v^2 R This force opposes the given force Net F =F- Bd^2v^2 R =RF- Bd^2v^2 R Net acceleration = RF-B^2d^2v mR 2. Velocity becomes constant when acceleration is 0. F m - B^2d^2v_0 mR =0 F m = B^2d^2v_0 mR _0= FR B^2d^2 3. Velocity at line t a=- dv dt _ 0 ^ v dv RF-l^2B^2v = _ 0 ^ t dt mR [l_n [RF-l^2B^2v ]1 -l^2B^2 ]_0^ v [ t Rm ]_0^ t [l_n (RF-l^2B^2v ) ]_ 0 ^ v = -tl^2B^2 Rm _n (RF-l^2B^2v )-ln(RF)= -t^2B^2t Rm 1- l^2B^2v Rf =e^ -l^2B^2t Rm l^2B^2v Rf =1-e^ -l^2B^2t Rm = FR l^2B^2 (1-e^ -l^2B^2v_0t Rv_0m )=v_0(1-e^ -Fv_0m )

The rectangular wire-frame, shown in figure, has a width d, mass …

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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 $t = 0.$

Find the acceleration of the frame when its speed has increased to $v.$
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 $v_0.$
Show that the velocity at time t is given by $\text{v}=\text{v}_0\Big(1-\text{e}^{-\frac{\text{ft}}{\text{mv}_0}}\Big)$

✓

Answer

emf developed $= Bdv ($when it attains a speed $v)$

Current $=\frac{\text{Bdv}}{\text{R}}$
Force $=\frac{\text{B}\text{d}^2\text{v}^2}{\text{R}}$
This force opposes the given force
Net F $=\text{F}-\frac{\text{B}\text{d}^2\text{v}^2}{\text{R}}=\text{RF}-\frac{\text{B}\text{d}^2\text{v}^2}{\text{R}}$
Net acceleration $=\frac{\text{RF}-\text{B}^2\text{d}^2\text{v}}{\text{mR}}$

Velocity becomes constant when acceleration is $0.$

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

Velocity at line $t$

$\text{a}=-\frac{\text{dv}}{\text{dt}}$
$\Rightarrow\int_{0}^{\text{v}}\frac{\text{dv}}{\text{RF}-\text{l}^2\text{B}^2\text{v}}=\int_{0}^{\text{t}}\frac{\text{dt}}{\text{mR}}$
$\Rightarrow\Big[\text{l}_\text{n}\big[\text{RF}-\text{l}^2\text{B}^2\text{v}\big]\frac{1}{-\text{l}^2\text{B}^2}\Big]_0^{\text{v}} \ \Big[\frac{\text{t}}{\text{Rm}}\Big]_0^{\text{t}} $
$\Rightarrow\Big[\text{l}_\text{n}\big(\text{RF}-\text{l}^2\text{B}^2\text{v}\big)\Big]_{0}^{\text{v}}=\frac{-\text{tl}^2\text{B}^2}{\text{Rm}}$
$\Rightarrow\text{l}_\text{n}\big(\text{RF}-\text{l}^2\text{B}^2\text{v}\big)-\text{ln}(\text{RF})=\frac{-\text{t}^2\text{B}^2\text{t}}{\text{Rm}}$
$\Rightarrow1-\frac{\text{l}^2\text{B}^2\text{v}}{\text{Rf}}=\text{e}^{\frac {-\text{l}^2\text{B}^2\text{t}}{\text{Rm}}}$
$\Rightarrow\frac{\text{l}^2\text{B}^2\text{v}}{\text{Rf}}=1-\text{e}^{\frac {-\text{l}^2\text{B}^2\text{t}}{\text{Rm}}}$
$\Rightarrow\text{v}=\frac{\text{FR}}{\text{l}^2\text{B}^2}\Big(1-\text{e}^{\frac{-\text{l}^2\text{B}^2\text{v}_0\text{t}}{\text{Rv}_0\text{m}}}\Big)=\text{v}_0(1-\text{e}^{-\text{Fv}_0\text{m}})$

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