College Physics
posted by Anand on .
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!

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
(42)i + (1+19)k + (12+1)k
= 2i +20j +12k
The force would be 1.6*10^19*23.4 N 
I am still not understanding it the way you explained it but your answer was correct.

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 
For the determinant, I get
(42)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