# A Charged Particle moving in Uniform Magnetic Field

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1. Aug 25, 2016

### Milind Chakraborty

1. The problem statement, all variables and given/known data
Problem given in the image attached.

Uniform Magnetic Field : B
Positive Charge : q
Uniform Velocity : v
Mass : m

Charged particle enters the magnetic field making an angle θ with the plane perpendicular to the magnetic field.

Width of the region of Magnetic field : d
d < (mv/qB)

2. Relevant equations
qvB = (mv2)/r

Where 'r' is the radius of the circular arc that the particle will move along in the magnetic field.

3. The attempt at a solution
All that I could see through is that in the given problem,

radius of the circular arc is greater than the width of the magnetic region and that the particle would come out through the other end of the region and move along the tangent at that point.

#### Attached Files:

• ###### PJ2SPYL137JEeBNBLTX9YtXE.jpg
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2. Aug 26, 2016

### BvU

Hello Milind,

here at PF we have a peculiar culture of letting the students do the work and helping them on the way. So you will need to take the first step -- see the guidelines . Not too difficult in this case: draw a chunk of circle to extend the straight line and start some analytic geometry work... .

You only have symbols to work with, but that's OK. It allows you to choose d << mv/qB and d = mv/qB as limiting cases.

(: and this latter situation is cumbersome for $\theta < {\pi\over 2}$ so you trust the exercise composer wasn't looking out carefully enough, but did intend to avoid such a complication. In short: it comes out on the right. Who knows you can earn extra brownie points for - carefully - pointing out this error
[edit2]: the Lorentz force is pointing ... which way ? )

And keep an eye on part ii) .

Last edited: Aug 26, 2016
3. Aug 26, 2016

### andrevdh

The angle theta is with respect to the plane perpendicular to the field.
This means that the particle enters the field with a speed v cos theta perpendicular to the width of the region where the field excists.