Hall Effect Derivation: Applications and Solved Examples

The Hall effect is the formation of a potential difference, also known as Hall voltage across an electrical conductor that is transverse to an electric current in the conductor and perpendicular to an applied magnetic field perpendicular to the current. 

• This effect was found by Edwin Hall in 1879. 
• The ratio of the induced electric field to the product of the current density and the applied magnetic field is defined as the Hall coefficient. 
• It is a property of the material used to make the conductor since its value is determined by the type, quantity, and qualities of the charge carriers that make up the current.

Key Terms: Electrical conductors, Hall voltage, Potential difference, Current density, Semiconductors, Electric field, Hall effect, Electric current

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Hall Effect

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In the presence of a magnetic field When current flows through a conductor the charge carriers, which are electrons, experience a transverse force that pushes them to one side of the conductor.

• This results in an excess positive charge on the opposite side of the conductor. 
• A potential difference between the sides of the conductor is formed due to the accumulation of charge carriers on one side of the conductor. 
• The existence of this measurable potential difference is known as the Hall effect.

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Hall Effect Derivation

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When a magnetic field B is applied to a conductor carrying current in the positive Y direction, the Lorentz Force on the electrons is given by

F_l = – Bev_d

Where

• e is the magnitude of the charge
• v_d is the drift velocity

Hall Effect

A positive charge builds up on one side due to the Lorentz force exerted in the negative Y direction, causing a negative charge buildup on the other side due to the Hall Effect.

As a result, an electric field is formed, which is given by

F_e = – eE_h

Where

• E_h is the electric field generated due to the Hall effect
• F_e is the force due to the electric field

In a neutral situation, Lorentz force and electric force are balanced. Therefore

F_l = F_e 

⇒ - Bev_d = – eE_h

⇒ E_h = Bv_d ….(i)

Let b be the length of the cross-section of the conductor, then the electric field generated due to the Hall effect is given by

E_h = V_h / b

⇒ V_h = E_hb

Where V_h is the Hall voltage.

Using equation (i), we get

V_h = Bv_db ….(ii)

Let J be the current density, then it is given by

J = – n_ee x v_d

⇒ v_d = - J / n_ee

Substituting the above expression in equation (i), we get

V_h = - BbJ / n_ee

Let d be the width of the cross-section of the conductor, then the current density is given by

J = I/A = I/db

⇒ V_h = - BbI / n_eedb

⇒ V_h = - BI / n_eed

This is the expression for the Hall voltage that develops across the conductor due to the Hall effect.

Also, we have V_h = E_hb

⇒ E_h = V_h/b

⇒ E_h = - BbJ / n_eeb

⇒ E_h = - BJ / n_ee

The above expression is the formula for the electric field generated due to the Hall effect, known as the Hall field.

From the above equation, we get

E_h / BJ = 1/n_ee

The ratio between the Hall field (E_h) to the product of the magnetic field (B) and current density is known as the Hall coefficient (R_h). Therefore

R_h = 1/n_ee

When the number of electrons exceeds the number of holes, the Hall coefficient formula for semiconductors becomes negative.

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Hall Effect Derivation in Semiconductors

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Because electrons and holes contribute to differing concentrations and mobilities in semiconductors, it is difficult to explain the Hall coefficient given above. 

As a result, the Hall coefficient is as follows for a simple explanation of a moderate magnetic field:

R_h = \(\frac{p \mu_H^2 - n\mu_e^2}{e(p\mu_H+n\mu_e)}\)

Where

• p is the hole concentration
• n is the electron concentration
• μ_H is the mobility of holes
• μ_e is the mobility of electrons
• e is the elementary charge

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Applications of Hall Effect

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The following are the applications of the Hall effect

• It determines whether a particular material is a semiconductor or an insulator.
• It is used in a magnetometer to measure the magnetic field.
• They are used in position sensing because they are resistant to water, mud, dust, and grime.
• They are used as Hall effect sensors in integrated circuits.
• It detects wheel speed and assists the anti-lock braking system accordingly.

Read More:

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Solved Examples

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Ques. Calculate the Hall voltage when the magnetic field is 8 A/m, the current is 4 A, the width is 5 m, and the concentration of carrier is 10²⁰

Ans. Given

• Magnetic field, B = 8 A/m
• Current, I = 4 A
• Width, d = 5 m
• Carrier concentration, n_e = 10²⁰

The formula for the Hall voltage is given by

V_h = - BI / n_eed

⇒ V_h = (8 x 4) / (10²⁰ x 1.6 x 10^-19 x 5)

⇒ V_h = 32/80 = 0.4 V

Ques. Calculate the width of the conductor slab having the concentration of the carrier is 10²¹ if the Hall voltage of 0.6 V develops across the slab when a magnetic field of 7 A/m is applied and the current flowing through the conductor is 3 A. 

Ans. Given

• Magnetic field, B = 7 A/m
• Current, I = 3 A
• Hall voltage, V_h = 0.6 V
• Carrier concentration, ne = 10²¹

The formula for the Hall voltage is given by

V_h = BI / n_eed

⇒ Width, d = BI / n_eeV_h

⇒ d = (7 x 3) / (10²¹ x 1.6 x 10^-19 x 0.6)

⇒ d = 21/96 = 0.21 m

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Things to Remember

• The Hall effect is the creation of a potential difference, sometimes known as a Hall voltage, across an electrical conductor that is perpendicular to the magnetic field is applied and transverse to the wire's electric current. 
• The Hall coefficient is referred to as the ratio of the induced electric field to the product of the current density and the applied magnetic field.
• Hall coefficient (R_h) is given by R_h = 1/n_ee
• The Lorentz force is applied to a charged particle traveling through a magnetic field B and an electric field E at a speed v.
• The formula for the Hall voltage is given by V_h = - BI / n_eed

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