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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: v x =(E/B){1-cos(qBt/m)} v y =(E/B)sin(qBt/m)


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)


Grade:upto college level

1 Answers

ROSHAN MUJEEB
askIITians Faculty 829 Points
one year ago

​=EE​orBB​;i^=v0​v0​​;k^=v0​Bv0​×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​=ay​t=mqE​T(i)

The charged particle under the action of magnetic field describes a circle in x-z plane (perpendicular toB) with:
T=Bq2πm​orω=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​=v0​cosωt=v0​cos(mqB​t)(ii)
vz​=v0​sinωt=v0​sin(mqB​t)(iii)

From Eqs. (i), (ii), and (iii)
v=vx​i^+vy​j^​+vz​k^
∴v=v0​cos(mqB​t)(v0​v0​​)+mqE​t(EE​)+v0​sin(mqB​t)(v0​Bv0​×B​)
Orv=cos(mqB​t)(v0​)+(mq​t)(E)+sin(mqB​t)(Bv0​×B​)
The path of the particle will be a helix of increasing pitch. The axis of the helix will be along y-axis.

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