# College Physics

posted by 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?

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

• College Physics - ,

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

• College Physics - ,

I am still not understanding it the way you explained it but your answer was correct.

• College Physics - ,

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 - ,

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