At a given instant, a particle with mass 5.00*10^-3kg and charge 3.50*10^-8C has

Question

At a given instant, a particle with mass 5.00*10^-3kg and charge 3.50*10^-8C has velocity of magnitude of 2.00*10^5m/s in the +y direction. It is moving in a uniform magnetic field of magnitude 0.8T directed 40 degrees clockwise from the -x direction. What are the magnitude and direction of the magnetic force on the particle?

I have found that the F magnitude of the magnetic force on the charge is 4.29*10^-3. However, I need help with the question below. I understand that it should be the x axis component of the F, but maybe someone can better explain to me what that means? Thank you.

To check your answer, you should calculate the magnitude of the force using the second formula: F=qvB?. What is the magnitude of B??

Explanation / Answer

Yes, that is the x component

Along the x direction, that is the cos of 40 degrees

Bx = (.8)(cos 40) = .6128 T

Then, for the check...

F = qvBx = (3.5 X 10-8)(2 X 105)(.6128) = 4.29 X 10-3 N so that works out...

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