Home / Magnetic Field and Magnetic Force Problem and Solutions

The Hall Effect (Crossed Fields) Problems and Solutions

Post a Comment

Basic Principle of the Hall Effect

When a conductor carries current II, electrons move with a drift velocity inside the material. If a magnetic field BB is applied perpendicular to the current, the moving charges experience a magnetic force known as the Lorentz force.

The Lorentz force is given by

$F=qvB$

where $q$= charge of the particle, $v$ = velocity of the charge and $B$ = magnetic field strength

This force pushes the electrons toward one side of the conductor. The accumulation of electrons creates an electric field inside the material.

Eventually, equilibrium is reached when the electric force balances the magnetic force.

At equilibrium:

$qE=qvB$

which leads to the velocity relation.

$v=\frac{E}{B}$

 This condition is commonly referred to as crossed electric and magnetic fields.

Hall Voltage Formula

The separation of charge inside the conductor produces the Hall voltage $V_H$ across the material.

The Hall voltage is given by:

$V_H=\frac{IB}{nqt}$

where $I$ = current in the conductor, $B$ = magnetic field strength, $n$ = charge carrier density, $q$= charge of the carrier and $t$= thickness of the conductor

This equation allows physicists to determine the number of charge carriers inside a material.

Problem#1

Figure 1 shows a portion of a silver ribbon with $z_1 = 11.8$ mm and $y_1 = 0.23$ 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. If there are $5.85 \times 10^{28}$ 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.

Fig.1

Answer:
Known:
Thickness of silver ribbon, $y_1 = 0.23 \ mm = 0.23 \times 10^{-3} \ m$
wide of silver ribbon, $z_1 = 11.8 \ mm = 11.8 \times 10^{-3} m$
a current of I = 120 A
Uniform magnetic field, B = 0.95 T
Charge carriers per unit volume of copper = $5.85 \times 10^{28} \ electrons/m^3$

(a) the magnitude of the drift velocity of the electrons in the x-direction is

$v_d = \frac{I}{y_1z_2nq}$

$v_d = \frac{3.0 \ A}{[0.23 \times 10^{-3} \ m \times 11.8 \times 10^{-3} \ m \times 5.85 \times 10^{28} \ electrons/m^3 \times 1.6 \times 10^{-19} \ C]}$

$v_d = 4.7 \times 10^{-3} \ m/s$

(b) the magnitude and direction of the electric field in the z-direction due to the Hall effect is

$v_d = \frac{E}{B}$

then,

$E = Bvd = 0.95 T \times 4.7 \times 10^{-3} \ m/s = 4.5 \times 10^{-3} \ N/C$, in the +z-direction.

(c) the Hall emf is potential difference between the two edges of the strip (at z = 0 and z = $z_1$) that result from the electric field calculated in part (b)

$\epsilon_{hall} = Ez_1 = (4.5 \times 10^{-3} \ N/C)(11.8 \times 10^{-3} \ m) = 5.3 \times 10^{-5} \ V$

Problem#2
Let Fig. 1 represent a strip of an unknown metal of the same dimensions as those of the silver ribbon in Problem#1. When the magnetic field is 2.29 T and the current is 78.0 A, the Hall emf is found to be 131 μV. What does the simplified model of the Hall effected in this section give for the density of free electrons in the unknown metal?

Answer:

Known:
a current of I = 78.0 A

Uniform magnetic field, B = 2.29 T

emf is V = 131 μV = $131 \times 10^{-6} \ V$

Area $A = y_1z_1$

the electric field, $E = \frac{\epsilon}{z_1}$ and │q│ = e.

The number density of charge carriers are measured by

$n = \frac{ByIz_1}{Aqε} = \frac{BzI}{y_1qε}$

$n=\frac{78.0 \ A \times 2.29 \ T}{2.3 \times 10^{-4} \ m \times 1.6 \times 10^{-19} \ C \times 1.31 \times 10^{-6} \ V}$

$n = 3.7 \times 10^{28} \ electrons/m^3$

Problem#3
In a Hall-effect experiment, a current of 3.0 A sent length wise through a conductor 1.0 cm wide, 4.0 cm long, and 10 mm thick produces a transverse (across the width) Hall potential difference of 10 μV when a magnetic field of 1.5 T is passed perpendicularly through the thickness of the conductor. From these data, find 
(a) the drift velocity of the charge carriers and 
(b) the number density of charge carriers. 
(c) Show on a diagram the polarity of the Hall potential difference with assumed current and magnetic field directions, assuming also that the charge carriers are electrons.

Answer:
Known:
a current of I = 3.0 A

wide of conductor $d = 1.0 \ cm = 1.0 \times 10^{-2} \ m$
length of conductor, $L = 4.0 \ cm = 4.0 \times 10^{-2} \ m$
Thickness of conductor, $t = 10 \ \mu m = 1.0 \times 10^{-6} \ m$
Uniform magnetic field, B = 1.5 T
Hall potential difference of $V = 10 \ \mu V = 10 \times 10^{-5} \ V$

(a) the drift velocity of the charge carriers is

$v_d = \frac{E}{B}$

with E is the electric field and can be written in term of the potential difference as;

$E= \frac{V}{d}$ 

Then,
$v_d = \frac{V}{dB} = \frac{10 \times 10^{-5} \ V}{1.0 \times 10^{-2} \ m \times 1.5 \ T}$

$v_d = 6.7 \times 10^{-4} \ m/s$

(b) the number density of charge carriers is

$n = \frac{Bi}{Vte} = \frac{I}{etdvd}$

$n=\frac{3.0 \ A}{1.6 \times 10^{-19} \ C \times 10 \times 10^{-6} \ m \times 1.0 \times 10^{-2} \ m \times 6.7 \times 10^{-4} \ m/s}$

$n = 2.8 \times 10^{29} \ /m^3$

(c) Diagram the polarity of the Hall potential difference with assumed current and magnetic field directions, assuming also that the charge carriers are electrons shown in the figure 2.
Fig.2
Problem#4
A strip of copper 150 mm thick and 4.5 mm wide is placed in a uniform magnetic field B of magnitude 0.65 T, with B perpendicular to the strip. A current i = 23 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 volume for copper is $8.47 \times 10^{28} \ electrons/m^3$.)

Answer:
Known:
Thickness of copper strip, $t = 150 \ \mu m = 150 \times 10^{-6} \ m$
Width of copper strip, $w = 4.5 \ mm = 4.5 \times 10^{-3} \ m$
Uniform magnetic field, B = 0.65 T
Current in the strip, i = 23 A
Charge carriers per unit volume of copper $= 8.47 \times 10^{28} \ electrons/m^3$

The number density of charge carriers are measured by

$n = \frac{Bi}{Vte}$

Then, potential difference is

$n = \frac{Bi}{Vte}$

$V = \frac{0.65 T \times 23 \ A}{8.47 \times 10^{28} \ electrons/m^3 \times 1.6 \times 10^{-19} \ C \times 150 \times 10^{-6} \ m}$

$V = 7.35 \times 10^{-6} \ V = 7.35 \ \mu V$

Problem#5
A metal strip 6.50 cm long, 0.850 cm wide, and 0.760 mm thick moves with constant velocity v through a uniform magnetic field B = 1.20 mT directed perpendicular to the strip, as shown in Fig. 3. A potential difference of 3.90 μV is measured between points x and y across the strip. Calculate the speed v.


Fig.3

Known:
Thickness of metal strip, $t = 0.760 \ mm = 0.76 \times 10^{-3} \ m$
Width of metal strip, $w = 0.850 cm = 0.85 \times 10^{-2} \ m$
Uniform magnetic field, $B = 1.20 \ mT = 1.20 \times 10^{-3} \ T$
A potential difference of $V = 3.90 \mu V = 3.90 \times 10^{-6} \ V$

For a free charge q inside the metal strip wirh velocity we have

$F = q\textbf{E} + v \times \textbf{B}$

We set this force equal to zero and use relation between (uniform) electric field and potential difference.

Then, the speed v is

$v=\frac{E}{B}$

with E is the electric field and can be written in term of the potential difference as;

$E=\frac{V_x-V_y}{Bdx}$

Hence,

$v = \frac{3.90 \times 10^{-6} \ V}{(1.20 \times 10^{-3} \ T \times 0.850 \times 10^{-2} \ m}$

$v = 0.382 \ m/s$ 

Problem#6

A rectangular conductor carries a current of 5 A in a magnetic field of 0.4 T. The charge carrier density is $3 \times 10^{28} \ m^{-3}$ and the thickness of the conductor is and the thickness of the conductor is . Find the Hall voltage.

Answer

The Hall voltage formula:

$V_H=\frac{IB}{nqt}$

Convert thickness: $t = 1 \ mm = 1 \times 10^{-3}$ m

Substitute the values:

$V_H=\frac{(5)(0.4)}{(6 \times 10^{28})(1.6 \times 10^{-19})(1 \times 10^{-3})}$

Thus, the Hall voltage is approximately

$V_H=2.1 \times 10^{-7}$ V

Problem#7

In a region of space, a uniform electric field and magnetic field are perpendicular to each other. The electric field is $E= 2000 \ V/m$. The magnetic field is $B=0.5 \ T$. Find the velocity of a particle that passes through the region undeflected.

Answer

For undeflected motion, the electric force must balance the magnetic force:

$qE = qvB$

Simplifying:

$v=\frac{E}{B}$

Substitute the values:

$v=\frac{2000 \ V/m}{0.5 \ T}=4.0 \times 10^3 \ m/s$

Thus, the particle moves with velocity

$v=4.0 \times 10^3 \ m/s$

Applications of the Hall Effect

Post a Comment for "The Hall Effect (Crossed Fields) Problems and Solutions"