Question

A strip of copper$\mathbf{\text{150}}$ m thick and $\mathbf{\text{4.5}}$ mm wide is placed in a uniform magnetic field$\stackrel{\mathbf{\rightharpoonup }}{\mathbf{B}}$of magnitude $\mathbf{\text{0.65}}$T, with$\stackrel{\rightharpoonup }{B}$ perpendicular to the strip. A currentlocalid="1663949700654" $\mathbf{\text{i=23}}$ A is then sent through the strip such that a Hall potential difference Vappears across the width of the strip. Calculate V. (The number of charge carriers per unit volume for copper islocalid="1663949722414" ${\mathbf{\text{8.47×10}}}^{\mathbf{\text{28}}}$electrons/m³.)

Short answer

The hall voltage is ${\text{V}}_{\text{H}}=\text{7.4}\times {\text{10}}^{\text{-6}}\text{V}$.

Step-by-step solution

Given

$\text{d}=\text{150µm}\left(\frac{{10}^{-6}\text{\hspace{0.17em}m}}{1\text{\hspace{0.17em}µm\hspace{0.17em}}}\right)=1.50\times {10}^{-4}\text{\hspace{0.17em}m}$

$\text{w}=\text{4.5mm}\left(\frac{{10}^{-3}\text{\hspace{0.17em}m}}{1\text{\hspace{0.17em}mm\hspace{0.17em}}}\right)=4.5\times {10}^{-3}\text{\hspace{0.17em}m}$

$\text{B}=\text{0.65T}$.

Determining the concept

The Hall Effect is the production of a voltage difference (the Hall voltage) across an electrical conductor, transverse to an electric current in the conductor, and an applied magnetic field perpendicular to the current.

Formulae are as follows:

${\text{F}}_{\text{m}}={\text{ev}}_{\text{d}}\text{B}$

${\text{v}}_{\text{d}}=\frac{\text{I}}{\text{neA}}$

${\text{F}}_{\text{e}}=\frac{{\text{V}}_{\text{H}}\text{ed}}{\text{A}}$

Where V_H is hall voltage, d is thickness, A is the area, V_dis drift velocity, F_e is electric force, I is current, e is the charge on particle, F_m is a magnetic force, and B is the magnetic field.

Determining the hall voltage

To find Hall voltage$\left({\text{V}}_{\text{H}}\right)$:

Here, both forces balance each other.

Hence,

${\text{F}}_{\text{m}}={\text{F}}_{\text{e}}$

${\text{ev}}_{\text{d}}\text{B}=\frac{{\text{V}}_{\text{H}}\text{ed}}{\text{A}}$

${\text{V}}_{\text{H}}=\frac{{\text{Av}}_{\text{d}}\text{B}}{\text{d}}$

Now, putting the formula of drift velocity,

$\begin{array}{c}{\text{V}}_{\text{H}}=\frac{\text{AIB}}{\text{neAd}}\\ =\frac{\text{IB}}{\text{ned}}\end{array}$

$\begin{array}{c}{V}_H=\frac{23A\times 0.65T}{8.47\times {10}^{28}{\text{\hspace{0.17em}m}}^{\text{-3}}\times 1.50\times {10}^{-6}m\times 1.6\times {10}^{-19}C}\\ =7.4\times {10}^{-6}V\end{array}$

Hence, the hall voltage is${\text{V}}_{\text{H}}=\text{7.4}\times {\text{10}}^{\text{-6}}\text{V}$.

Therefore, by using the concept of hall voltage, the hall voltage through the copper strip can be determined