# Two particles in a magnetic field

1. Feb 25, 2005

### sony

Illustration:
http://home.no.net/erfr1/images/2.jpg
"A particle, X with mass m and charge q starts with an inital velocity zero, and follows the red path indicated on the drawing. It follows a half circle when entering the magnetic field.

A new particle, Y with mass m(same) and charge 2q starts from the same place as X. Is the following true og untrue:

1. The force on Y is double the force on X in the electric field.
2. The force on Y is double the force on X in the magnetic field.
3. The time Y uses on the half circle is half the time X uses.

1. Fex=qE and Fey=2qE. So that's true.
2. Fmx=qvB = q*sqrt(qU/m) * B.
And Fmy = qvb = 2q * sqrt(2qU/m) * B. So this statement is false.
3. I don't get this.

It's half a circle, so v=Pi*r/T
And I have tried the formula: v^2/r = 4Pi^2*r / T^2

But everything I use involves either v, or r og both. And the radius of Y has to be smaller right? Since the mag. force on Y is the biggest. But I don't know the radius! And the speed is such an ugly expression, and I always end up with the square root of an expression, so the difference can't be 1/2.

The book says that statement 3 is true...

Thanks

Last edited by a moderator: Apr 21, 2017
2. Feb 25, 2005

### dextercioby

$$T_{X}=\frac{\pi r_{X}}{v_{x}}$$ (1)
$$T_{Y}=\frac{\pi r_{Y}}{v_{Y}}$$ (2)
$$r=\frac{mv}{qB}$$ (3)

Use (3) for the 2 particles and then combine with (1) & (2) to reach the result.

Daniel.

3. Feb 25, 2005

### sony

Aaaah! Thank you. So the speed is ruled out and I get:
Tx= Pi*m/qB
Ty=1/2 * Pi*m/qB
Ty=1/2Tx
I can't believe I spent so much time on this :). Stupid me.

Thanks!