# Relating Electric Field and Magnetic Field

PeachBanana

## Homework Statement

Protons move in a circle of radius 7.70cm in a 0.536 T magnetic field. What value of electric field could make their paths straight?

Unsure

## The Attempt at a Solution

r = 0.0770 m
B = 0.536 T
m = 1.6*10^-27 kg
q = 1.6*10^-19 C

E = F / q

I'm not sure how to get force without a velocity.

E_M_C
Hi PeachBanana,

Consider the answers to these questions:

If both an electric and magnetic field are present, how can you write the net force on the proton?

In order for the proton to move in a straight line, the net force on the proton must be what?

What kind of velocity does the proton have when it's circling around in a magnetic field? And what is the equation for that kind of velocity?

This should give you a hint:
http://en.wikipedia.org/wiki/Cyclotron_motion

Saitama
What should be the direction and magnitude of electric field so that protons go in straight direction?
If the force due to electric field is opposite to the direction of force and equal in magnitude to force exerted by the magnetic field, the protons go in a straight path.
i.e qvB=qE.

To get the velocity, i guess you are given sufficient data, what is the relation between velocity and radius?

PeachBanana
I have now come up with:

qvB = mv^2 /r

qb = mv/r

v = rqB/m

v = (0.0770 m)(1.9*10^-19)(0.536 T) / (1.67 * 10^-27 kg)

v = 4.0*10^6 m/s

Hmm.

PeachBanana
Just realized I solved for the wrong variable completely. Since F = qvb and I know "v," I can solve for "F." When I know "F", I can then use the fact that E = F/q.

E_M_C
Since F = qvb and I know "v," I can solve for "F." When I know "F", I can then use the fact that E = F/q.

Edit: Ah yes, I see what you're doing. That should give you the correct numerical answer. Just make sure you specify the directions of the fields. For future reference, you should look at the Lorentz Force. This is the net force on a charged particle when both electric and magnetic fields are present.

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