A straight segment of a current-carrying wire has a current element I L where I

Question

A straight segment of a current-carrying wire has a current element I L where I = 8.89 A and L = [ Lx i + Ly j + Lz k ], where i, j, and k are unit vectors pointing in the x, y, and z directions, respectively. The values for Lx, Ly, and Lz are -0.81 m, -0.07 m, and -0.43 m, respectively. The segment is in a region with a uniform magnetic field given by B = [ Bx i + By j + Bz k ], where the components of B are ( 4.100E-2, 9.490E-1, and 2.290E-1) T, respectively. Find the components of the force on the segment of wire.

a) What is Fx?

b) What is Fy?

c) What is Fz?

Explanation / Answer

The force on a current carrying wire is

F=ILxB

So Lets just calculate LxB (since in your case L is the vector) and multiply the result by I.

LxB = (LyBz-LzBy)i + (LzBx-LxBz)j + (LxBy-LyBx)k

LxB = ((-0.07)(2.290E-1)-(-0.43)(9.490E-1))i + ((-0.43)(4.100E-2)-(-0.81)(2.290E-1))j + ((-0.81)(9.490E-1)-(-0.07)(4.100E-2))k

LxB = 0.39204i +0.16786j -0.76582k

Now we need to multiply all of these by I:

F = 3.485i + 1.49j - 6.81k, All in N, since your L was m, I was A and B was T.

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