# The Force on a Current in a Magnetic Field

## Homework Statement

The x, y, and z components of a magnetic field are Bx = 0.10 T, By = 0.13 T, and Bz = 0.17 T. A 25 cm wire is oriented along the z axis and carries a current of 4.5 A. What is the magnitude of the magnetic force that acts on this wire?

## Homework Equations

F=ILB (sin theta)

## The Attempt at a Solution

Well, I have
F= (4.5A)(25e-2 m)(B?) (sin 90)

I don't know which value for B to use or how to incorporate the three values together. I tried using the three B values to find three different force values, but I don't know how to "add" the three together.

This may be totally wrong but when I plugged Bx, By and Bz into the above equation I got
Fx=.01125N
Fy=.01462N and
Fz=.01913N

...but I'm pretty sure this is wrong

The B values are components of a vector. How do you get the length of a vector from its components? The theta in your equation is the angle between the conductor and the magnetic field and is not 90 degrees. How would you compute it?

Well, if I only had x and y values I would use the pythagoreom theorum(sp) to add the two, but I don't know how to add in the third z component.

As for determining theta I'm thinking that it would be -tan y/x
soo -tan .13/.10 = 52.4 degrees
but what about the z axis?
I'm so confused.

The z axis is what is throwing me totally off. I don't know how what to do with it.

There's a pythagorean theorem in three dimensions as well. Take the square root of the sum of all of the components squared. Along the z-axis in three dimensions means it's direction vector can be written as (0,0,1). The easy way to find the angle between two vectors is to use the dot product. cos(theta)=a.b/(|a|*|b|). Does that ring a bell?

I have the same problem, but with different numbers...

"The x, y, and z components of a magnetic field are Bx = 0.12 T, By = 0.12 T, and Bz = 0.17 T. A 25 cm wire is oriented along the z axis and carries a current of 4.6 A. What is the magnitude of the magnetic force that acts on this wire?"

I can not figure it out with what was discussed in earlier posts