A charged particle of mass m and positive charge q moves in a uniform constant e

Question

A charged particle of mass m and positive charge q moves in a uniform constant electric and magnetic fields, E pointing in the y direction and B in the z direction (crossed E and B fields). Suppose the particle is initially at the origin and has initial velocity Vxo in the x direction.

a) Find the electric force on the particle.

b) Find the magnetic force on the particle.

c) The net electric and magnetic force on the particle is called the Lorentz force. Using Newton?s second law, write down equations for the acceleration of the particle due to the Lorentz force in all three directions. Your acceleration should be in terms of the velocity, charge, electric and magnetic fields.

d) Prove that there is a unique value of Vxo, called the drift speed Vdr, for which the particle moves undeflected through the fields. Find an expression for Vdr in terms of E and B.

Explanation / Answer

1. force between two charges is givrn by coulombs law, as F = kq1q2/r^2

i.e when two charges are separated by a disatnce on r, /

attraction or repulsion forces between charges takes place as upon on one charge there acts the electric field given by the second charge

so in F -= kq1q2/r62

Kq2/r^2 ius called the eletrioc field and is denoted by E

so F = Eq.

that is when ever a charged particle is placed in the EF, that is acted upon by eletrical force

2. similarly when ever a charegd particle is introduced in the region os mgnetic field, this charge is acted upon by magnetic force

this force is directly poroportional to magnetidue of charge, q

directly proportion to velocity of charge v

and directly prop to magneti9c field B

so magnetic force F = qvB

magnetic field acts only if charged pa[rticle us in motiion, if it is at rest no mag force is acts on it

( this is differennt from that of electric field)

3. sum of these two called Lorentz force = Fe + FB

= qE + q(Vx B)

4. from the concepts of em theory,

using Fradays law of induction,

int E.ds = dphiB/dt

phi B = h dE = Bh dx

so dE/dx = -dB/dt

dE/dx = -kEx cos (kx-wt)

dB/dt = -w Bm(cos (kx-wt)

KEm cos (Kx-wt) = wBx cos (kx-wt)

w/K = v

so Em/Bm = Vd

or Vd = E/B

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