Relating Electric Field and Magnetic Field

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SUMMARY

The discussion focuses on determining the electric field required to make protons move in a straight line while they are in a 0.536 T magnetic field and moving in a circular path with a radius of 7.70 cm. The relevant equations include the relationship between electric and magnetic forces, specifically qvB = qE, where q is the charge of the proton, v is its velocity, and B is the magnetic field strength. The calculated velocity of the protons is 4.0 x 106 m/s, which is derived from the equation v = rqB/m. The net force acting on the protons must be zero for them to travel in a straight line, necessitating that the electric force equals the magnetic force in magnitude but opposite in direction.

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PeachBanana
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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?

Homework Equations



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.
 
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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
 
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?
 
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.
 
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.
 
PeachBanana said:
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.
 
Last edited:

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