X−Y plane shown in the figure contains uniform magnetic field B→=Bk→ for y0. A particle having charge q and mass m travels along y-axis. At origin of co-ordinate system velocity of particle is v0 and it entres the region containing magnetic field. Assume that particle is subjected to a frictional force f→=−αv→ i.e. frictional force is proportional to velocity. Assume frictional force is large enough so that particle remains inside region y0 at all times. The only force acting on particle are frictional force and magnetic force. Particle will remain in x−y plane as no magnetic force will act along z-axis. So F→=−αv→+qv→×B→ . The x-coordinate where particle comes to rest is given by λqBmV0α2+(qB)2 . Find λ .

Question

X−Y  plane shown in the figure contains uniform magnetic field  B→=Bk→  for  y>0.  A particle having charge q and mass m travels along y-axis. At origin of co-ordinate system velocity of particle is v0  and it entres the region containing magnetic field. Assume that particle is subjected to a frictional force f→=−αv→   i.e. frictional force is proportional to velocity. Assume frictional force is large enough so that particle remains inside region  y>0  at all times. The only force acting on particle are frictional force and magnetic force. Particle will remain in  x−y  plane as no magnetic force will act along  z-axis. So F→=−αv→+qv→×B→ . The x-coordinate where particle comes to rest is given by λqBmV0α2+(qB)2 . Find λ  .

Accepted Answer

Using above equation it can easily shown that $m \frac{d^{2} x}{{dt}^{2}} = - α \frac{dx}{dt} + qB \frac{dy}{dt}$ and $m \frac{d^{2} y}{{dt}^{2}} = - α \frac{dy}{dt} - qB \frac{dx}{dt}$
If we integrate above equations from the time particle enters the region $y > 0$ to the time particle comes to rest, then we will get, $m Δ V_{x} = - α Δ x + q B Δ y {mΔV}_{y} = - α Δy - qBΔx$ . Solve the above equations we get $x = \frac{q B m V}{α^{2} + {(q B)}^{2}}$

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