# Homework Help: Hard Magnetic Field Problem.

1. Nov 27, 2007

### PFStudent

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

14. Figure 28-36 shows a metallic block, with its faces parallel to coordinate axes. The block is in a uniform magnetic field of magnitude ${0.020{{.}}T}$. One edge length of the block is $25{{.}}cm$; the block is not drawn to scale. The block is moved at $3.0{{.}}m/s$ parallel to each axis, in turn, and the resulting potential difference $V$ that appears across the block is measured. With the motion parallel to the y-axis, $V = 12{{.}}mV$; with the motion parallel to the z-axis, $V = 18{{.}}mV$; with the motion parallel to the x-axis, $V = 0{{.}}V$. What are the block lengths
(a) ${d_{x}}$,
(b) ${d_{y}}$, and
(c) ${d_{z}}$?

[Fig. 28-36]

2. Relevant equations

$${\vec{E}_{P1}} = {{\frac{k_{e}q_{1}}{{r_{P1}}^2}{\hat{r}_{1P}}}}$$

$${V_{P1}} = {\frac{k_{e}q_{1}}{{r_{P1}}}}$$

$${\vec{F}_{E}} = {q{\vec{E}}}$$

$${\Delta{V}_{p}} = {{-}{\int_{r_{0}}^{r_{1}}}{\vec{E}_{p1}(r)}{\cdot}{d{\vec{r}}}}$$

$${\vec{E}} = {{-}{\nabla}{V(r)}}$$

$${\vec{E}} = {{-}{\frac{\partial}{\partial{r}}}{\left[{V(r)}\right]}{\hat{r}}}$$

$${\vec{F}_{B}} = {q{\vec{v}}{\times}{\vec{B}}}$$

3. The attempt at a solution

Yea, this problem does not appear to be easy at all. What I first did was assume that since the block is being moved along each axis at the same velocity. Then the potential difference between the points along the axes where the block stops moving with respect to the point at the origin, should be the same for all three points (at least I think so).

After thinking about the above for a while, I thought it was really not getting me anywhere. Then I began to just consider the movement of the block along a single axes. Considering the y-axis first (as I am assuming they are moving the block along each axis in the order they list the axes (ie. origin to y, y to z, and z to x)). Then I have,

origin to y:

$${\Delta{V}_{y}} = {{-}{\int_{y_{0}}^{y_{1}}}{\vec{E}_{p1}(y)}{\cdot}{d{\vec{y}}}}$$

Since, the block begins to move from the origin,

${y_{0}} = 0{{.}}m$

So, I now have,

$${\Delta{V}_{y}} = {{-}{|\vec{E}_{}(y)|}{|y_{1}|}{cos{\theta}_{y}}}$$

However, it is from here that I am not sure what to do next, I think I'm on the right track. I'm just not sure how to proceed from here. I am guessing that I have to re-express ${cos{\theta}_{y}}$ in terms of: ${y_{1}}$ and ${z_{1}}$, however I'm not quite sure how I should do that.

Any help is appreciated.

Thanks,

-PFStudent

Last edited by a moderator: May 3, 2017
2. Nov 28, 2007

### PFStudent

Hey,

I'm still looking over this problem and I figured I'm on the right track where I left off.

So, if I continue working through this problem--then for the next two parts I get the following,

$${\Delta{V}_{z}} = {{-}{\int_{z_{0}}^{z_{1}}}{\vec{E}_{p1}(z)}{\cdot}{d{\vec{z}}}}$$

Where I get,

$${\Delta{V}_{z}} = {{-}{|\vec{E}_{}(z)|}{z_{1}}{cos{\theta}_{z}}}$$

Where,

$${{\theta}_{z}} = 180^{o}$$

So, now I have,

$${\Delta{V}_{z}} = {{|\vec{E}_{}(z)|}{z_{1}}}$$

But, I still do not see how to solve for the dimensions of the box.

How do I solve for the dimensions of the box?

Any help is appreciated.

Thanks,

-PFStudent.

3. Nov 30, 2007

### PFStudent

Hey,

So, any ideas on how to solve for the dimensions of the block?

Thanks,

-PFStudent