# Magnetic fields and motor principle

• Mar 29th 2008, 10:28 AM
johett
Magnetic fields and motor principle
1) An electron moves at a speed of 5.0x10^6m/s perpedicular to a uniform magnetic field. The path of the electron is a circle of radius 1.0x10^-3m.

a)what is the magnitude of the magnetic field?

b)what is the magnitude of the electron's acceleraton in the field

2) A beam of protons move in a circle of radius 0.22m at right angles to a 0.35T magnetic field.

a)What is the speed of each proton?

b)what is the magnitude of the centripetal force acting on each proton.

Tried a few equations but still can't seem to get the right answers....
• Mar 29th 2008, 04:44 PM
topsquark
Quote:

Originally Posted by johett
1) An electron moves at a speed of 5.0x10^6m/s perpedicular to a uniform magnetic field. The path of the electron is a circle of radius 1.0x10^-3m.

a)what is the magnitude of the magnetic field?

b)what is the magnitude of the electron's acceleraton in the field

The electron is moving at a constant speed in a circular path. So applying Newton's second law in the radial direction (+r toward the center) at a particular instant, we get
$\displaystyle \sum F_r = F_{mag} = F_C = \frac{mv^2}{r}$
where
$\displaystyle F_{mag} = qvB$

So
$\displaystyle qvB = \frac{mv^2}{r}$

$\displaystyle B = \frac{mv}{qr}$
(This is the "cyclotron formula" rewritten a bit.)

To get the acceleration of the electron, notice that the acceleration will simply be the centripetal acceleration, so
$\displaystyle a = \frac{v^2}{r}$

Problem 2 is almost identical to this one.

-Dan
• Mar 30th 2008, 06:37 AM
johett
$\displaystyle B= \frac{mv}{qvr}$

$\displaystyle = (9.11x10^-31)(5.0x10^6m/s) / (1.602x10^-19)(5.0x10^6m/s)(1.0x10^-3)$
$\displaystyle =5.0x10^-9$

I still can't seem to get the right answer... I used 1.602x10^-19 for q and 9.11x10^-31 for the mass of the electron. The answer is suppose to be 2.8x10^-2 T.
• Mar 30th 2008, 09:52 AM
Flippy
Topsquark made a little mistake, which you should have noticed when entering the formula in your calculator.

The correct formula for B should be:
$\displaystyle B= \frac{mv}{qr}$

This will yield $\displaystyle B = 0,0284 T$ or $\displaystyle B = 2,84 \times 10^{-2} T$
• Mar 30th 2008, 10:19 AM
topsquark
Quote:

Originally Posted by Flippy
Topsquark made a little mistake, which you should have noticed when entering the formula in your calculator.

The correct formula for B should be:
$\displaystyle B= \frac{mv}{qr}$

This will yield $\displaystyle B = 0,0284 T$ or $\displaystyle B = 2,84 \times 10^{-2} T$

Thanks for the catch. I have fixed it in my original post.

-Dan