## Sunday, March 04, 2007

### Charged Particles in Magnetic and Electric Fields

Here is a multiple choice question which appeared in AIEEE 2002 question paper:
If an electron and a proton having same momenta enter perpendicular to a magnetic field, then
(a) curved path of electron and proton will be same (ignoring the sense of revolution)
(b) they will move undeflected
(c) curved path of electron is more curved than that of the proton
(d) path of proton is more curved

The radius of the circular path of the electron is obtained by equating the magnetic force to the centripetal force: qvB = mv²/r. The radius ‘r’ is therefore given by r = mv/qB. The radius is therefore directly proportional to the momentum (mv) and inversely proportional to the charge (q) of the particle.
The momenta are given as equal in the problem. Since the proton and the electron have the same charge magnitudes and are moving in the same magnetic field B, they will follow paths of the same radius[Option (a)].
You might have noted that the path of a charged particle in an electric field is generally parabolic. This is because of the fact that in a uniform electric field, the electric force on the particle has the same direction everywhere. The motion is similar to the projectile motion in a gravitational field. Now, consider the following question:
A particle of charge ‘+q’ and mass ‘m’ is projected with a velocity ‘v’at an angle ‘θ’ with respect to the horizontal, in an electric field ‘E’ which is directed vertically downwards. If there are no gravitational or magnetic fields, the horizontal range of the particle is
(a) (v²sin2θ)/E (b) (v²sin2θ)/qE (c) (mv²sin2θ)/E (d) (mv²sin2θ)/qE (e) (qmv²sin2θ)/E
In the case of the motion of a projectile in a gravitational field, the expression for horizontal range is R = (v² sin2θ)/g. In the present case, the gravitaional acceleration ‘g’ is replaced by the acceleration produced by the electric field.
Acceleration produced by the electric field = Force/ Mass = qE/m. The correct option therefore is (d).
The following MCQ appeared in AIIMS 2004 question paper:
The cyclotron frequency of an electron gyrating in a magnetic field of 1 T is approximately
(a) 28 MHz (b) 280 MHz (c) 2.8 GHz (d) 28 GHz
The cyclotron frequency is the frequency with which a charged particle describes circular path in a magnetic field and is given by f =qB/2πm with usual notations. [You can get it this way: qvB = mrω² where ω is the angular frequency. Substituting v = ωr in this, we get ω = qB/m. Frquency f = ω/2π =qB/2πm].
Substituting for the mass and charge of the electron, we have
f = (1.6×10–19 ×1)/ (2π×9.1×10–31 ) = 28×109 Hz = 28 GHz.

1. Anonymous9:27 PM

Usеful information. Luсky me I diѕсovered
youг site aсcidentally, аnd
I am ѕtunnеd why this accident dіdn't took place in advance! I bookmarked it.
Also visit my website - Fish Oil

2. Anonymous2:49 AM

Hey there! Do you know if they make any plugins to help with Search Engine Optimization?
I'm trying to get my blog to rank for some targeted keywords but
I'm not seeing very good gains. If you know of any please
share. Thanks!

Here is my webpage :: terry bandy

3. Anonymous6:10 PM

Appreciating the hard work you put into your blog
and detailed information you offer. It's good to come across a blog every once in a
while that isn't the same old rehashed information. Great read!