What is the Force on a Proton in a 0.5 T Magnetic Field?

AI Thread Summary
The discussion focuses on calculating the magnetic force on a proton in a 0.5 T magnetic field, with the proton traveling at 3.4 × 10^5 m/s at an angle of 28 degrees. The formula used is F = vBq(sin θ), leading to a calculated force of 1.27 × 10^-14 N. However, this result differs from the textbook answer of 2.7 × 10^13 N, prompting questions about the charge value used for the proton and the clarity of the proton's direction of travel. The participants emphasize the importance of correctly identifying the angle and the magnetic field's orientation. Accurate calculations are essential for understanding the magnetic force on charged particles.
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Homework Statement



A proton is projected into a magnetic field of 0.5 T directed into the page. If the proton is traveling at 3.4 × 10^5 m/s in a direction [up 28 ̊ right], what is the magnitude and direc- tion of the magnetic force on the proton?

Homework Equations



F = vBq(sin θ)

The Attempt at a Solution



(3.4 * 10^5)*(.5)(1.6*10^-19)(sin 28) = 1.27 * 10^-14

The book says the answer is 2.7 * 10^13

Maybe the charge for the proton is wrong.
 
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I am assuming that the magnetic field is perpendicular into the page.
The direction of travel of the proton is not so clear.
 
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