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

(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 -
**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

(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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