Posted by Anand on Saturday, March 10, 2012 at 1:48am.
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!

College Physics  drwls, Saturday, March 10, 2012 at 1:59am
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

College Physics  Anand, Saturday, March 10, 2012 at 2:04am
I am still not understanding it the way you explained it but your answer was correct.

College Physics  drwls, Saturday, March 10, 2012 at 3:46am
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

College Physics  drwls, Saturday, March 10, 2012 at 8:17am
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
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