Magnetism> In the figure of the charge q is released...

In the figure of the charge q is released at the origin then find its x and y components of velocity as a function of time t.Given:Electric field E acts along the +ve y axis and in xy plane there exists a uniform magnetic field B acting along +ve z axis. Mass of charged particle is m.Ans:vx=(E/B){1-cos(qBt/m)}vy=(E/B)sin(qBt/m)

Amit Saxena , 11 Years ago

Grade upto college level

1 Answers

ROSHAN MUJEEB

Last Activity: 4 Years ago

=EEorBB;i^=v0v0;k^=v0Bv0×B0

Force due to electric field will be along y-axis. Magnetic force will not affect the motion of charged particle in the direction of electric field (or y-axis) so.
ay=mFe=mqE,vy=ayt=mqET(i)

The charged particle under the action of magnetic field describes a circle in x-z plane (perpendicular toB) with:
T=Bq2πmorω=T2π=mqB

Initially(t=0)velocity was along x-axis. Therefore, magnetic force(Fm)will be along positive z-axis[Fm=q(v0×B)].
Let it this force makes an angleθwith x-axis at time t, thenθ=ωt.
∴vx=v0cosωt=v0cos(mqBt)(ii)
vz=v0sinωt=v0sin(mqBt)(iii)

From Eqs. (i), (ii), and (iii)
v=vxi^+vyj^+vzk^
∴v=v0cos(mqBt)(v0v0)+mqEt(EE)+v0sin(mqBt)(v0Bv0×B)
Orv=cos(mqBt)(v0)+(mqt)(E)+sin(mqBt)(Bv0×B)
The path of the particle will be a helix of increasing pitch. The axis of the helix will be along y-axis.