Asked by Anand

A proton moves with a velocity of vector v = (6ihat - 1jhat + khat) m/s in a region in which the magnetic field is vector B = (ihat + 2jhat - 3khat) T. What is the magnitude of the magnetic force this charge experiences?

answer in Newtons.

I tried F=qvb but the vectors are just stumping me. Any help is welcome!

Answers

Answered by drwls
Compute the determinant:
|i j k|
|6 -1 1|
|1 2 -3|
Then calculate the magnitude of the resulting vector. Then multiply that by e.

Why do you have an unnecessary 1 in front of the velocity j component? Is that a typing error?

For the determinant, I get
(4-2)i + (1+19)k + (12+1)k
= 2i +20j +12k

The force would be 1.6*10^-19*23.4 N
Answered by Anand
I am still not understanding it the way you explained it but your answer was correct.
Answered by drwls
You need to learn the method of taking the cross product of two vectors with determinants. That is what I did.

For a brief summary see
http://www.ucl.ac.uk/mathematics/geomath/level2/mat/mat121.html
Answered by drwls
For the determinant, I get
(4-2)i + (1+19)j + (12+1)k
(a typo error was corrected)
= 2i +20j +12k

The magnitude of that vector is
sqrt[2^2 + 20^2 + 12^2] = sqrt548 = 23.4

The force would be 1.6*10^-19*23.4 N
= 3.74*10^-18 N
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