21.9 HALL EFFECT
21.9.1 Hall Effect
1. Hall effect
Consider a conductor or semiconductor of rectangular cross-section in which a constant current flows. A uniform magnetic field is applied to the conductor in a direction which is perpendicular to the direction of the current, as shown in Figure 21.51(a). Then an emf is developed across points P and Q on the conductor, where PQ is perpendicular to the direction of the magnetic field and the direction of the current. This production of an emf in a current-carrying conductor or semiconductor due to the application of magnetic field in a direction which is perpendicular to the direction of the current is known as the Hall effect.
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| Figure 21.51 (a) Hall effect in a conductor showing the development of Hall voltage across points P and Q when a magnetic field is applied perpendicular to the current. (b) Magnetic force acts vertically upwards on moving electrons due to the magnetic field directed into the page. |
2. Reasons for getting Hall effect
(a) When a conduction electron in the conductor drifts to the left through the magnetic field, a magnetic force will act on it.
(b) Since the field points into the page and is perpendicular to the direction of the current, the magnetic force will have to point vertically upwards, as shown in Figure 21.51(b).
(c) Because of this, all those conduction electrons that move through the magnetic field tend to drift upwards to the region at point P. As a result, there is an increase in conduction electrons in the upper region, thus causing point P to become negatively charged.
(d) The drifting of a large number of conduction electrons towards P causes the lower region at point Q to be depleted of conduction electrons. This causes the electric potential at point Q to be less than that at P.
(e) A difference in electric potential exists between points P and Q. Hence, a voltage is produced across PQ.
3. Hall voltage
The constant voltage developed across PQ is known as the Hall voltage. Equation for the Hall voltage:
(a) Since the upper surface at P is negatively charged and the lower surface at Q is positively charged, an electric field is produced in the conductor between P and Q.
(b) Now a conduction electron moving through the magnetic field is not only acted on by a magnetic force Fm but also by an electric force Fe. Fe acts upwards on the electron while Fm acts downwards.
(c) Initially, Fm is greater than Fe. Hence, a net force which points upwards acts on a conduction electron, causing it to drift to the upper surface. However, the upper surface becomes more and more negatively charged as more and more conduction electrons continue to drift upwards. At the same time, the lower surface becomes more and more positively charged. The voltage across PQ increases.
(d) The strength of the electric field increases, hence increasing the magnitude of the electric force acting on an electron. Finally, Fe becomes equal in magnitude to Fm. Now an electron entering the two fields would drift through the conductor without being deviated to the upper or lower surface.
(e) Under this condition we have
Fe = Fm
eE = Bev sin 90°
where E, B and v represent the electric field strength, magnetic field strength and the drift speed of the electron respectively. But we have
E = VH / d
where VH and d represent the Hall voltage and the distance between the upper and lower surfaces respectively.
VH = Bvd
But the drift speed is given by
v = I / (neaA)
where I is the current flowing in the conductor or semiconductor
n is the number of conduction electrons per unit volume
A is the cross-sectional area of the conductor
We have
A = ad
where a is the thickness of the conductor
VH = B ( I / (nead) ) d
VH = BI / (nea)
EXAMPLE 21.25
A strip of semiconductor is 1.0 mm thick. A current of 10 mA flows through the semiconductor. A magnetic field of strength 500 mT is then applied to the semiconductor in a direction which is parallel to the side of 1.0 mm thickness and perpendicular to the direction of the current. Determine the Hall voltage across the breadth of the semiconductor.
(Electron charge = −1.6 × 10−19 C; number of charge carriers per unit volume = 1.7 × 1022 m−3)
Answer
VH = BI / nea = (500 × 10−3)(10 × 10−3) / (1.7 × 1022)(1.6 × 10−19)(1.0 × 10−3) = 1.8 mV
21.9.2 Uses of Hall Effect
Uses of Hall effect:
1. We can use the equation
VH = Bdv
to measure the drift speed v of the charge carriers if VH, B, d can be measured.
2. We can use the equation
VH = BI / nea
to measure
(a) the number of charge carriers per unit volume n if VH, B, I, a can be measured
(b) the magnetic field strength B.
3. We can determine the sign of the charge carried by the charge carriers that move in the conductor or semiconductor. If the charge carriers are
(a) negatively charged, the potential at P would be negative relative to Q
(b) positively charged, the potential at P would be positive relative to Q.
EXAMPLE 21.26
A current of 0.25 A flows in a strip of material which is 0.20 mm thick and 5.0 mm wide. The strip is placed inside a uniform magnetic field of strength 200 mT, with the larger surface of the strip perpendicular to the direction of the field. The Hall voltage developed across the two opposite sides of the strip is 15 μV. Determine
(a) the number of charge carriers per unit volume for the conductor
(b) the drift speed of the charge carriers.
Answer
(a)
n = BI / VHea
= (200 × 10−3)(0.25) / (15 × 10−6)(1.6 × 10−19)(0.2 × 10−3)
= 1.0 × 1026 m−3
(b)
v = VH / Bd
= 15 × 10−6 / (200 × 10−3)(5.0 × 10−3)
= 1.5 × 10−2 m s−1
21.9.3 Determination of Magnetic Field Strength Using Hall Voltage
1. An instrument for measuring magnetic field strength
The Hall voltage VH produced by a magnetic field of strength B applied to a strip of current-carrying conductor or semiconductor is given by
VH = ( I / nea ) B . . . . (1)
Suppose that we replace the magnetic field with another field whose strength B0 is known but the quantities I, n, e and a remain unchanged. Let the new Hall voltage produced by this field be VHO. Then we have
VHO = ( I / nea ) B0 . . . . (2)
(1) ÷ (2),
VH / VHO = B / B0
or,
B = ( B0 / VHO ) VH
= k VH
where the value of k is known as B0 is known and VHO can be measured. Since k is known for the strip of conductor or semiconductor, we can use the strip as a component in an instrument to measure the unknown strength of any magnetic field. The scale of the voltmeter that is used to read the Hall voltage VH can be converted into a scale for reading the magnetic field strength B.
2. Preference for semiconductor strip
We have
VH = ( B / ea ) ( I / n )
where the current
I = p.d. / resistance
I ∝ 1 / resistance for a fixed p.d.
For a copper strip, we have
VHC ∝ 1 / ncρc
where nc ~ 1029 m−3, ρc ~ 10−8 Ω m.
For a silicone strip of similar size placed inside the same magnetic field, we have
VHS ∝ 1 / nsρs
where ns ~ 1015 m−3, ρs ~ 103 Ω m.
VHC = ( ns / nc ) ( ρs / ρc ) VHS
VHC = ( 1015 / 1029 ) ( 103 / 10−8 ) VHS
= 10−3 VHS
This means that the Hall voltage produced by a given magnetic field in a copper strip could be several hundred to a thousand times smaller than the Hall voltage produced by the same field in a semiconductor strip. Hence, it would be advantageous to use a semiconductor strip instead of a metallic strip if we intend to obtain a large Hall voltage.
EXAMPLE 21.27
A magnetic field of strength 200 mT produces a Hall voltage of 15 μV in a current-carrying strip placed appropriately inside the field. When the field is replaced by another, the Hall voltage becomes 25 μV. Determine the strength of the second field for the same magnetic field.
Answer
B = kVH
200 mT = k(15 μV)
Bnew = k(25 μV)
Bnew = (25 / 15)(200)
= 333 mT
Applications of Hall Effect
| 1. Measuring Magnetic Field Strength | The Hall effect is widely used to measure the strength of a magnetic field. By measuring the Hall voltage, the value of the magnetic field can be determined accurately. This principle is used in devices known as Hall probes. |
| 2. Identifying Type of Charge Carriers | The Hall effect can determine whether the charge carriers in a conductor or semiconductor are positive or negative. If the Hall voltage indicates a negative potential, the charge carriers are electrons; if positive, they are holes. |
| 3. Measuring Charge Carrier Density | It is used to calculate the number of charge carriers per unit volume in a material. This is important in studying electrical properties of conductors and semiconductors. |
| 4. Current Sensors | Hall effect sensors are used to measure electric current without direct contact. These sensors are commonly found in power supplies, battery systems, and electronic circuits. |
| 5. Position and Speed Sensors | The Hall effect is used in sensors to detect position, speed, and rotation. It is widely applied in automotive systems such as wheel speed sensors, crankshaft position sensors, and brushless DC motors. |
| 6. Proximity Sensors | Hall effect sensors are used in proximity sensing devices to detect the presence of nearby magnetic objects without physical contact. |
Conclusion
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The Hall effect is an important concept in physics that explains how a voltage is produced across a conductor when it carries current in the presence of a magnetic field. This phenomenon helps us understand the behavior of charge carriers in conductors and semiconductors.
Through the Hall effect, we can determine key physical quantities such as magnetic field strength, charge carrier density, and drift velocity. It also allows us to identify whether the charge carriers are positive or negative. In modern technology, the Hall effect plays a crucial role in the development of sensors used in various applications, including current measurement, position detection, and magnetic field sensing. Overall, the Hall effect is not only a fundamental theory in electromagnetism but also a highly practical tool widely used in science and engineering. |
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