- #1
wjdgone
<Moderator's note: Moved from a technical forum and therefore no template.>
A proton (q = 1.60 x 10-19) is in a uniform, 0.500 T magnetic field. This proton has velocity components vx = 1.50 x 105, vy = 0, and vz = 2.00 x 105 m/s. Find the force on the proton at t=0.
2. Homework Equations
F= qv x B
I was wondering specifically about how to go about calculating the v in F = qv x B. Now, I understand the method using vectors (F = q(vxi + vzk) x Bi = qvzBj).
But this method isn't my first instinct when solving this problem. I'd want to go about graphing the velocity components, draw a parallelogram, and use a2 + b2 = c2 to avoid using the component method. What I got using this method was v = 2.50 x 105, instead of the 2.00 x 105 that you get when using the first, component method above. Plugging 2.50 x 105, I got F = 2.00 x 10-14, compared to 1.60 x 10-14 using the first method.
Is this second method legit? Why is there a difference in the results?
Homework Statement
A proton (q = 1.60 x 10-19) is in a uniform, 0.500 T magnetic field. This proton has velocity components vx = 1.50 x 105, vy = 0, and vz = 2.00 x 105 m/s. Find the force on the proton at t=0.
2. Homework Equations
F= qv x B
The Attempt at a Solution
I was wondering specifically about how to go about calculating the v in F = qv x B. Now, I understand the method using vectors (F = q(vxi + vzk) x Bi = qvzBj).
But this method isn't my first instinct when solving this problem. I'd want to go about graphing the velocity components, draw a parallelogram, and use a2 + b2 = c2 to avoid using the component method. What I got using this method was v = 2.50 x 105, instead of the 2.00 x 105 that you get when using the first, component method above. Plugging 2.50 x 105, I got F = 2.00 x 10-14, compared to 1.60 x 10-14 using the first method.
Is this second method legit? Why is there a difference in the results?
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