# Force exerted by a magnetic field

## 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.

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 dont 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
Homework Helper
Gold Member
I rearranged the equation so B=F/(IxL). I plugged all of the numbers in but I dont 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:

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

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 Bx, By, and Bz.

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