Question

Figure E27.49 shows a portion of a silver ribbon with z₁ = 11.8 mm and y₁ = 0.23 mm, carrying a current of 120 A in the +x-direction. The ribbon lies in a uniform magnetic field, in the y-direction, with magnitude 0.95 T. Apply the simplified model of the Hall effect presented in Section 27.9. If there are 5.85 * 10²⁸ free electrons per cubic meter, find (a) the magnitude of the drift velocity of the electrons in the x-direction; (b) the magnitude and direction of the electric field in the z-direction due to the Hall effect; (c) the Hall emf

Short answer

1. The drift velocity of the electrons in the x-direction is${\mathbf{v}}_{\mathbf{d}}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{7}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{m}\mathbf{/}\mathbf{s}$
2. The magnitude and direction of the electric field in the z-direction due to the Hall effect is ${\mathbf{E}}_{\mathbf{z}}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{V}\mathbf{/}\mathbf{m}$
3. The hall emf${\mathbf{E}}_{\mathbf{Hall}}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{3}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{5}}\mathbf{V}$

Step-by-step solution

Drift velocity

The drift velocity is given by

${\mathbf{V}}_{\mathbf{d}}\mathbf{=}\frac{{\mathbf{J}}_{\mathbf{x}}}{\mathbf{n}\left|q\right|}$

Determine the drift velocity

(a)

The drift velocity is given by

${\mathbf{V}}_{\mathbf{d}}\mathbf{=}\frac{{\mathbf{J}}_{\mathbf{x}}}{\mathbf{n}\left|q\right|}$

Where,$\mathbf{n}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{85}\mathbf{\times }{\mathbf{10}}^{\mathbf{28}}{\mathbf{m}}^{\mathbf{-}\mathbf{3}}$

The drift velocity is

$\begin{array}{r}{\mathbf{J}}_{\mathbf{x}}\mathbf{=}\frac{\mathbf{I}}{\mathbf{A}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\\ \mathbf{=}\frac{\mathbf{I}}{{\mathbf{y}}_{\mathbf{1}}{\mathbf{z}}_{\mathbf{1}}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\\ \mathbf{=}\frac{\mathbf{120}\mathbf{A}}{\left(0.23\times {10}^{-3}m\right)\mathbf{(}\mathbf{0}\mathbf{.0118}\mathbf{m}\mathbf{)}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\\ \mathbf{=}\mathbf{4}\mathbf{.42}\mathbf{\times }{\mathbf{10}}^{\mathbf{7}}\mathbf{A}\mathbf{/}{\mathbf{m}}^{\mathbf{2}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}}\\ \mathbf{And}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\\ {\mathbf{v}}_{\mathbf{d}}\mathbf{=}\frac{\mathbf{4}\mathbf{.42}\mathbf{\times }{\mathbf{10}}^{\mathbf{7}}\mathbf{A}\mathbf{/}{\mathbf{m}}^{\mathbf{2}}}{\left(5.85\times {10}^{28}{m}^{-3}\right)\left(1.602\times {10}^{-19}C\right)}\\ \mathbf{=}\mathbf{4}\mathbf{.7}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{m}\mathbf{/}\mathbf{s}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\end{array}$

Therefore, the drift velocity of the electrons in the x-direction is ${\mathit{v}}_{\mathbf{d}}\mathbf{=}\mathbf{4}\mathbf{.7}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{m}\mathbf{/}\mathbf{s}\mathbf{}\mathbf{}$

Determine the magnitude and direction of electric field

(b)

Magnetic force equals to the electric field

$\mathbf{\left|}\mathbf{q}\mathbf{\right|}{\mathbf{E}}_{\mathbf{z}}\mathbf{=}\mathbf{\left|}\mathbf{q}\mathbf{\right|}{\mathbf{v}}_{\mathbf{d}}{\mathbf{B}}_{\mathbf{y}}$

Or,

$\begin{array}{r}{\mathbf{E}}_{\mathbf{z}}\mathbf{=}{\mathbf{V}}_{\mathbf{d}}{\mathbf{B}}_{\mathbf{y}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\\ \mathbf{=}\left(4.7\times {10}^{-3}m/s\right)\mathbf{(}\mathbf{0}\mathbf{.95}\mathbf{T}\mathbf{)}\\ \mathbf{=}\mathbf{4}\mathbf{.5}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{V}\mathbf{/}\mathbf{m}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\end{array}$

Therefore, the magnitude and direction of the electric field in the z-direction due to the Hall effect is${\mathit{E}}_{\mathbf{z}}\mathbf{=}\mathbf{4}\mathbf{.5}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{V}\mathbf{/}\mathbf{m}\mathbf{}$

Determine the hall emf

(c)

The hall emf is

$\begin{array}{r}{\mathbf{E}}_{\mathbf{\text{Hall}}}\mathbf{=}{\mathbf{Ez}}_{\mathbf{1}}\mathbf{=}\left(4.5\times {10}^{-3}V/m\right)\mathbf{(}\mathbf{0}\mathbf{.0118}\mathbf{m}\mathbf{)}\\ \mathbf{=}\mathbf{5}\mathbf{.3}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{5}}\mathbf{V}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\end{array}$

Therefore, the hall emf is${\mathbf{E}}_{\mathbf{\text{Hall}}}\mathbf{=}\mathbf{5}\mathbf{.3}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{5}}\mathbf{V}\mathbf{}\mathbf{}$