# electric field and magnetic field - proton deflection

by scholio
Tags: deflection, electric, field, magnetic, proton
 P: 160 1. The problem statement, all variables and given/known data when protons travelling north in a horizontal plane enter a region of uniform magnetic field of 0.8Teslas in the downward direction, they are deflected into a horizontal circle of radius 0.2 meters. what is the magnitude and direction of a uniform electric field applied over the same region of space that will allow the protons to pass through the region undelflected 2. Relevant equations radius r = mv/qB where m is mass, v is velocity, q is charge, B is magnetic field electromagnetic force F = qE + qv X B where X indicates cross product electric field force F_E = qE where E is electric field magnetic force F_B = qv X B where B is magnetic field charge of electron/proton = 1.6*10^-19 coulombs mass proton m = 1.67*10^-27 kg 3. The attempt at a solution i used : since the proton must not be deflected, i assumed electric field force must equal magnetic force so: qE = qv X B i then used radius eq, r = mv/qB and solved for v --> v = qBr/m and subbed it in for v to get: qE = q(qBr/m) X B and solved for E --> E = (qBr/m)Bsin(theta), i have q = 1.6*10^-19 coulombs, B = 0.8 Teslas, r = 0.2 meters, mass m = 1.67*10^-27 kg but what is theta? is my approach correct? cheers
 PF Patron HW Helper P: 2,688 Your approach for finding the magnitude of the E field is correct. The direction of the B-field is given in the problem, though it is worded in a somewhat confusing way. They tell you that the protons are moving north and that the B field is pointing in the "downward" direction. I believe this means "into the plane of the page." Considering this to be the case, what is your angle?
 P: 160 which page? the page the problem is written on? in that case, the angle would be 90 degrees. correct?
P: 1,133

## electric field and magnetic field - proton deflection

If you were to take a path northward over the Earth's surface, southward would be the direction opposite of northward (which is antiparallel to the motion and so the proton would be unaffected), and downward would be the direction into the Earth (which is perpendicular to the motion and is what the question is probably referring to---so you're correct, angle is 90 degrees)...heh, question was worded kind of confusingly. There is a trick to determining the angle (if I'm right about this): the particle will move in a circle if the field is completely 90 degrees perpendicular to the motion...otherwise the path is helical.
PF Patron
HW Helper
P: 2,688
 Quote by scholio which page? the page the problem is written on? in that case, the angle would be 90 degrees. correct?
Yup. You got it.
 P: 160 thanks, so using theta = 90 degrees and E = (qBr/m)Bsin(theta), i have q = 1.6*10^-19 coulombs, B = 0.8 Teslas, r = 0.2 meters, mass m = 1.67*10^-27 kg i got electric field E = 1.23*10^7 coulombs/meter
 PF Patron HW Helper P: 2,688 Looks good. Good job!

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