# Magnetism, component values, and three dimensions

1. Jul 31, 2007

### exi

1. The problem statement, all variables and given/known data

A magnetic field has these components:

x: B = 0.056 T
y: B = 0.046 T

The particle has a charge of +6.5e-5 C, has a velocity of 2.6e3 m/s, and is moving along the z axis.

Questions:

1: What's the magnitude of the net magnetic force on the particle?

2: What's the angle the net force makes with respect to the x axis?

2. Relevant equations

F = qvb(sinΘ)
Component addition of the x- and y- values.

3. The attempt at a solution

I have the answer to #1 by adding the x- and y- components and getting a resulting magnitude of 0.0725 T at 39.4007° between those axes. (0.01225 N)

On #2, I apparently incorrectly assumed that the net force was directed along the z-axis and is therefore 90° from the x-axis. Is the angle I found earlier (39.4007°) indeed the answer to this question? I'm unsure.

2. Jul 31, 2007

### Gokul43201

Staff Emeritus
I think you got lucky with the correct magnitude because x^2 + y^2 = y^2 + x^2. I suspect the values you have for Fx and Fy and in fact reversed.

Use the Lorentz force equation in vector form (what do you know about the cross - or vector - product) to determine magnitudes and directions of each component.

3. Jul 31, 2007

### exi

The x- and y- magnitude components are correct; just double-checked them.

Not sure what you're getting at as far as a Lorentz equation; that's new to me.

4. Jul 31, 2007

### Gokul43201

Staff Emeritus
Okay, so did you use $F_x = -qv_zB_y$ and $F_y = qv_zB_x$?

This is simply the scalar breakup of the vector equation $\vec{F} = q (\vec{v} \times \vec{B})$. Have you seen this equation in this form? Have you dealt with cross products yet?

Last edited: Jul 31, 2007
5. Jul 31, 2007

### exi

Whoa, for the first part? No, I did it the good ol' fashioned way, with $|B| = \sqrt{0.056^2 + 0.046^2}$.

And yeah, that does look familiar, albeit in already-crossed $F_B = qvBsin\theta$ form. I think I'm having a bit of a hard time with the angle bit, since I have an angle from the x- and y- components and am not exactly sure what to do with the rest of what's here.

6. Jul 31, 2007

### Gokul43201

Staff Emeritus
Okay, in that case, all you need to remember is that the force, F is always perpendicular to both the field, B, and the velocity v. So, it must be one of the 2 normals to B in the x-y plane. Which one, is determined by the right hand rule.

7. Aug 1, 2007

### exi

If I do what I think is the RHR for this question (with my thumb pointing upwards, towards the ceiling, for the particle moving along the z-axis, and my fingers extending in the direction of the x-y plane magnetic field at 39.4007° to the right), and if the force is in the direction indicated by my palm, it feels like that would put it at 129° to x. Am I going about this correctly?

8. Aug 1, 2007

### Gokul43201

Staff Emeritus
Looks good.

9. Aug 1, 2007

### exi

I thought so too - but it was incorrect and my last submission on that question. Key won't be available until tomorrow am.

Wonder where I went wrong...

10. Aug 1, 2007

### Matthaeus_

$$\displaystyle \vec{B} = (56\,\mathrm{mT})\hat{\i} + (46\,\mathrm{mT})\hat{\j}$$

$$\displaystyle \vec{v} = (2.6\,\mathrm{km/s})\hat{k}$$

When you have components, it is often wasier to calculate the cross product by finding the determinant of the matrix:

$$\displaystyle \vec{v}\times\vec{B} = \begin{array}{|ccc|} \hat{\i} & \hat{\j} & \hat{k}\\ 0 & 0 & 2600\\ 0.056 & 0.046 & 0 \end{array} = (-119.6)\hat{\i} + (145.6)\hat{\j}$$

Now you multiply this by q and you have the force:

$$\displaystyle \vec{F} = (-q \cdot 119.6)\hat{\i} + (q\cdot 145.6)\hat{\j}$$.

The magnitude of a vector $$\vec{F} = a\hat{\i} + b\hat{\j}$$ is, as you know, $$||F|| = \sqrt{a^2 + b^2}$$ and the angle of the vector with the x axis is $$\theta = \arctan{\frac b a}$$.

11. Aug 1, 2007

### exi

Thanks for the explanation.