Force exerted by a magnetic field

[apt8]

Homework Statement

a 10cm long straight wire is parallel with the z axis and carries a current of 4.0A in the +z direction. The force on this wire due to a uniform magnetic field B is -0.20N i+0.20N j. If this wire is rotated so that it is parallel with the x-axis with the current in the +x direction, the force on the wire becomes 0.20N k. Find B.

Homework Equations

F=I(LxB)

The Attempt at a Solution

I rearranged the equation so B=F/(IxL). I plugged all of the numbers in but I don't understand how to do the dot product if I is not a vector. Can I use some kind of dot product between F and L?

 

[collinsmark]

I rearranged the equation so B=F/(IxL). I plugged all of the numbers in but I don't understand how to do the dot product if I is not a vector. Can I use some kind of dot product between F and L?

I think you mean cross product, but anyway...

I think you can solve this problem by getting more detailed with what the cross product actually is.

One of the equations you've listed is

\vec F=I(\vec L \times \vec B).

Being more explicit with the cross product, this is the same thing as:

\vec F = I\left( \left| <br /> \begin{tabular}{ l c r }<br /> i &amp; j &amp; k \\<br /> L_x &amp; L_y &amp; L_z \\<br /> B_x &amp; B_y &amp; B_z \\<br /> \end{tabular}<br /> \right| \right)

where the thing inside of the parenthesis represents the determinant of the matrix.

What makes this problem a lot easier, is that some of the L components are zero in the first case, and other components are zero in the second case. So even though it looks ugly now, it's actually pretty easy to generate a few equations to solve for B_x, B_y, and B_z.

 

[apt8]

Yes I did mean the cross product. Thank you very much this helped a lot!