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प्रश्न
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).
उत्तर
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 = `(Bdv)/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=(Bd^2v)/R`
As the force is in direction opposite to that of the motion of the frame.
Therefore, Net force is given by
Fnet = F - FB
`F_"net"=F-(Bd^2v^2)/R=(RF-Bd^2v)/R`
Applying Newton's second law
`(RF-Bd^2v)/R=ma`
Net acceleration is given by `a=(RF-Bd^2v)/(mR)`
(b) Velocity of the frame becomes constant when its acceleration becomes 0.
Let the velocity of the frame be v0
\[\frac{F}{m} - \frac{B^2 d^2 v_0}{mR} = 0\]
\[ \Rightarrow \frac{F}{m} = \frac{B^{{}_2} d^2 v_0}{mR}\]
\[ \Rightarrow v_0 = \frac{FR}{B^2 d^2}\]
As the speed thus calculated depends on F, R, B and 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
\[a = \frac{dv}{dt}\]
\[ \Rightarrow \frac{RF - d^2 B^2 v^2}{mR} = \frac{dv}{dt}\]
\[\frac{dv}{RF - d^2 B^2 v^2} = \frac{dt}{mR}\]
Integrating
\[ \Rightarrow \int\limits_0^v \frac{dv}{RF - d^2 B^2 v^2} = \int\limits_0^t \frac{dt}{mR}\]
\[ \Rightarrow \left[ \ln(RF - d^2 B^2 v) \right]_0^v = - d^2 B^2 \left[ \frac{t}{Rm} \right]_0^t \]
\[ \Rightarrow \ln(RF - d^2 B^2 v) - \ln (RF) = - \frac{d^2 B^2 t}{Rm}\]
\[ \Rightarrow \frac{d^2 B^2 v}{RF} = 1 - e^{- \frac{d^2 B^2 t}{Rm}} \]
\[v = \frac{FR}{l^2 B^2}\left( 1 - e^{- \frac{B^2 d^2 v_0 t}{R v_0 m}} \right)\]
\[v = v_0 (1 - e^{- Ft/ v_0 m} ) ............\left[ \because F = \frac{B^2 d^2 v_0}{R} \right]\]
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