Chapter 4: Q49E (page 915)

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

Expert verified

The drift velocity of the electrons in the x-direction is $v_{d} = 4.7 \times 10^{- 3} m / s$
The magnitude and direction of the electric field in the z-direction due to the Hall effect is $E_{z} = 4.5 \times 10^{- 3} V / m$
The hall emf $E_{Hall} = 5.3 \times 10^{- 5} V$

Step by step solution

Drift velocity

The drift velocity is given by

$V_{d} = \frac{J_{x}}{n | q |}$

Determine the drift velocity

(a)

The drift velocity is given by

$V_{d} = \frac{J_{x}}{n | q |}$

Where, $n = 5.85 \times 10^{28} m^{- 3}$

The drift velocity is

$\begin{aligned} J_{x} = \frac{I}{A} \\ = \frac{I}{y_{1} z_{1}} \\ = \frac{120 A}{(0 .23 \times 10^{- 3} m) (0 .0118 m)} \\ = 4 .42 \times 10^{7} A / m^{2} \\ And \\ v_{d} = \frac{4 .42 \times 10^{7} A / m^{2}}{(5 .85 \times 10^{28} m^{- 3}) (1 .602 \times 10^{- 19} C)} \\ = 4 .7 \times 10^{- 3} m / s \end{aligned}$

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

Determine the magnitude and direction of electric field

(b)

Magnetic force equals to the electric field

$| q | E_{z} = | q | v_{d} B_{y}$

Or,

$\begin{aligned} E_{z} = V_{d} B_{y} \\ = (4 .7 \times 10^{- 3} m / s) (0 .95 T) \\ = 4 .5 \times 10^{- 3} V / m \end{aligned}$

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

Determine the hall emf

(c)

The hall emf is

$\begin{aligned} E_{Hall} = {Ez}_{1} = (4 .5 \times 10^{- 3} V / m) (0 .0118 m) \\ = 5 .3 \times 10^{- 5} V \end{aligned}$

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

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Short Answer

Step by step solution

Drift velocity

Determine the drift velocity

Determine the magnitude and direction of electric field

Determine the hall emf

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