How Do You Calculate the Hall Potential Difference in a Copper Strip?

[PFStudent]

Homework Statement

12. A strip of copper ${150{\textcolor[rgb]{1.00,1.00,1.00}{.}}{\mu}m}$ thick and ${4.5{\textcolor[rgb]{1.00,1.00,1.00}{.}}mm}$ wide is placed in a uniform magnetic field ${\vec{B}}$ of magnitude ${0.65{\textcolor[rgb]{1.00,1.00,1.00}{.}}T}$, with ${\vec{B}}$ perpendicular to the strip. A current ${i = 23{\textcolor[rgb]{1.00,1.00,1.00}{.}}A}$ is then sent through the strip such that a Hall potential difference $V$ appears across the width of the strip.

Calculate $$V$$. (The number of charge carriers per unit volulme for copper is $$8.47{\times}{10^{28}}$$ electrons/$$m^{3}$$).

Homework Equations

$$q = n_{e}e, {\textcolor[rgb]{1.00,1.00,1.00}{.}}{\textcolor[rgb]{1.00,1.00,1.00}{.}}{\textcolor[rgb]{1.00,1.00,1.00}{.}}{n_{e}} = \pm1, \pm 2, \pm 3,...,$$

e $\equiv$ elementary charge

$$e = 1.60217646 {\times} 10^{-19}{\textcolor[rgb]{1.00,1.00,1.00}{.}}C$$

$${n_{e}} = {\pm}N, {\textcolor[rgb]{1.00,1.00,1.00}{.}}{\textcolor[rgb]{1.00,1.00,1.00}{.}}{\textcolor[rgb]{1.00,1.00,1.00}{.}}{N} = 1, 2, 3,...,$$

$${N_{V}} = \frac{n_{e}}{V}$$

$${n_{e}} = {N_{V}}{V}$$

$${N_{V}} = {\frac{BI}{{\Delta{V}}{le}}}$$

Where, $${\Delta{V}}$$ is the Hall potential difference.

The Attempt at a Solution

This seems like a straight forward problem, here is how I worked through it.

$${\Delta{V}} = {\frac{BI}{{N_{V}}{le}}}$$

Let $I$ = current, ${\Delta}{V} = V$, and $\left({\frac{N}{V}}\right)$ be the number of charge carriers per unit volume. So, since we're dealing with electrons,

$${N_{V}} = -{\left({\frac{N}{V}}\right)}$$

Where,

${\left({\frac{N}{V}}\right)} = {{8.47}{\times}{10^{28}}}$ electrons/${m^{3}}$

However, since they gave me two distances: thickness (${t}$) and width (${w}$); which one is ${l}$?

I thought at first, it was the width because isn't that how the Hall potential difference is defined, as the potential across the width of a strip?

The book used the thickness, so I am wondering why?

Any help is appreciated.

Thanks,

-PFStudent

 

[Doc Al]

To understand why the formula for Hall voltage contains thickness (not width) you need to review its derivation. If your text isn't clear on the matter, check this out: http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/hall.html"

 

[PFStudent]

Hey,

To understand why the formula for Hall voltage contains thickness (not width) you need to review its derivation. If your text isn't clear on the matter, check this out: http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/hall.html"

Thanks for the link Doc Al.

So, I noticed they always used the thickness in their examples,..so in a nutshell is that how the Hall Potential is defined?

That is to say, in defining the Hall Potential--if we consider a thin strip (say like copper) of thickness ${t}$, width ${w}$, and length ${l}$; in a magnetic field ${\vec{B}}$ perpendicular to the face of the strip, carrying a current ${I}$ through one end of the strip and out the other.

Then there is a potential difference ${\Delta{V}}$ as a result of the electric force ${{\vec{F}}_{E}}$ and magnetic force ${{\vec{F}}_{B}}$ opposing each other due to the orientation of the two fields. This is a result of the magnetic field pushing the positive charges to one side of the thin strip. The placement of the positive charges on one side of the strip creates an electric field with the negative charges left on the other side of the strip. Resulting in the creation of a potential difference ${\Delta{V}}$ known as the Hall Potential Difference. Given as,

$${\Delta{V}} = {\frac{BI}{{N_{V}}{te}}}$$

Is that all right?

Thanks,

-PFStudent

 

[PFStudent]

Hey,

Yea, still a little unclear on this.

Any help would be appreciated.

Thanks,

-PFStudent

 

[rock.freak667]

The hall voltage is established when the resultant force exerted on the charge carriers is zero...that is when $F_E=F_B$

 

[PFStudent]

Hey,

The hall voltage is established when the resultant force exerted on the charge carriers is zero...that is when $F_E=F_B$

Ok, that makes sense but that does not quite explain why the Hall Potential is dependent on the thickness ${t}$, of the strip. Why is that?

Thanks,

-PFStudent

 

[rock.freak667]

Well it all comes from the derivation of the formula
start with
$$F_E=F_B$$
=> $$Ee=Bev$$
$$E=Bv$$
Re:$$E=\frac{V_H}{d}$$ where d= thickness

$$\frac{V_H}{d}=Bv$$
Re: $$v=\frac{I}{nAe}$$

$$\frac{V_H}{d}=\frac{BI}{nAe}$$
so that:

$$V_H=\frac{Bd}{nAe}$$
now A=td where l is the length of the material
$$V_H=\frac{Bd}{ntde}$$
$$V_H=\frac{B}{net}$$
so t is actually the length which is basically the width

 

[Doc Al]

Ok, that makes sense but that does not quite explain why the Hall Potential is dependent on the thickness ${t}$, of the strip. Why is that?

Sorry I didn't get back to this sooner. In a nutshell, while the Hall voltage is defined across the width of the strip, it only depends on the thickness of the strip not the width. If you check out the derivation on the link I gave, you'll see how it comes about.

Well it all comes from the derivation of the formula
start with
$$F_E=F_B$$
=> $$Ee=Bev$$
$$E=Bv$$

Good.

Re:$$E=\frac{V_H}{d}$$ where d= thickness

I'd say that:
$$E=\frac{V_H}{W}$$, where W = width, not thickness.

$$\frac{V_H}{d}=Bv$$
Re: $$v=\frac{I}{nAe}$$

$$\frac{V_H}{d}=\frac{BI}{nAe}$$
so that:

$$V_H=\frac{Bd}{nAe}$$
now A=td where l is the length of the material
$$V_H=\frac{Bd}{ntde}$$
$$V_H=\frac{B}{net}$$
so t is actually the length which is basically the width

I can't quite follow this. I'd rewrite it this way:

$$E = vB$$

$$\frac{V_H}{W} = vB$$
(where W is width)

$$v = \frac{I}{n e A}$$

where A is cross-sectional area = Width*thickness = Wt, so:
$$\frac{V_H}{W} = \frac{I B}{n e Wt}$$

Note how the width cancels out, leaving the dependence on thickness:
$$V_H = \frac{I B}{n e t}$$

 

[rock.freak667]

Well I guess width and thickness would depend on which way the electrons are flowing in the conducting material...as in my diagram..d as in the height of the conducting material.