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Test: Hall Effect - Electronics and Communication Engineering (ECE) MCQ


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10 Questions MCQ Test Electronic Devices - Test: Hall Effect

Test: Hall Effect for Electronics and Communication Engineering (ECE) 2025 is part of Electronic Devices preparation. The Test: Hall Effect questions and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus.The Test: Hall Effect MCQs are made for Electronics and Communication Engineering (ECE) 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Hall Effect below.
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Test: Hall Effect - Question 1

In Hall effect, a difference voltage is produced _______ to electric current in the conductor, and to an applied _____ field perpendicular to the current.

Detailed Solution for Test: Hall Effect - Question 1
  • Hall effect is the production of a voltage difference across an electrical conductor that is transverse to an electric current and to an applied magnetic field perpendicular to the current
  • When current flows in a semiconductor and magnetic field are applied perpendicularly, then the voltage is produced across the two side plates of the semiconductor bar, this is known as the Hall effect
  • The Voltage produced is known as Hall voltage

The following diagram resembles an experimental setup for the observation of the Hall effect

 

Here we could observe that 

  • Hall voltage VH is developed transverse to the current in the conductor
  • The hall voltage is also developed perpendicular to the applied magnetic field

Hall effect finds major applications in

  • Hall sensors
  • Used for proximity and displacement measurements
  • Used in RPM,speed measurements
Test: Hall Effect - Question 2

Hall effect is useful for the measurement of a semiconductor’s

Detailed Solution for Test: Hall Effect - Question 2

Hall Effect is basically the production of voltage (known as hall voltage) across conductor transverse to an electric current and to an applied magnetic field perpendicular to the current.
Application of Hall Effect:

  • To determine the type of semiconductor
  • To calculate the carrier concentration
  • To determine mobility (hall mobility)
  • To Measure magnetic flux density

From Hall Effect we have:

Where the hall coefficient is given as:

Hall Effect can be used to measure the carrier concentration and type of semiconductor.

Test: Hall Effect - Question 3

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

Detailed Solution for Test: Hall Effect - Question 3

Given Data:

• Magnetic field, B = 8 A/m

• Current, I = 4 A

• Width of the material, W = 5 m

• Carrier concentration, n = 10^20 m⁻³

• Charge of an electron, e = 1.6 × 10⁻¹⁹ C

Formula for Hall Voltage:

The Hall voltage is given by:

V_H = (B * I) / (n * e * W)

where:

• B is the magnetic field

• I is the current

• n is the carrier concentration

• e is the charge of an electron

• W is the width of the material

Step-by-Step Calculation:

1. Calculate charge density:

   ρ_c = n × e

   Substituting values:

   ρ_c = (10²⁰) × (1.6 × 10⁻¹⁹)

   ρ_c = 1.6 × 10¹ = 16

2. Calculate Hall Voltage:

   V_H = (B * I) / (n * e * W)

   Substituting values:

   V_H = (8 × 4) / (10²⁰ × 5 × 1.6 × 10⁻¹⁹)

   V_H = 32 / (5 × 1.6 × 10¹)

   V_H = 32 / 8 = 0.4 V

Final Answer:

V_H = 0.4 V

Test: Hall Effect - Question 4

For an n-type Ge specimen, width = 4 mm, length = 1 mm, current (along the length of specimen) = 1 mA, magnetic field (perpendicular to the current flow direction) = 0.1 Wb/m2 and Hall voltage magnitude = 0.005 V. Calculate the majority carriers density.

Detailed Solution for Test: Hall Effect - Question 4

Concept:
The hall voltage VA is given by:

ρc: Charge density
RH: Hall coefficient

W is the side across which the magnetic field is applied.
Calculation:
W = Thickness = 4 mm, I = 1 mA, VH = 0.005 V, B = 0.1 Wb/m2

Since hall voltage is positive, the majority charge carriers will be holes, with the concentration calculated as:

p ≈ 3 × 1019 m-3

Test: Hall Effect - Question 5

For a particular material, the Hall coefficient is found to be zero. The material is

Detailed Solution for Test: Hall Effect - Question 5

Hall coefficient is defined as:



We can find the electron concentration as:


ρc = charge density
σn = nμnq and σp = pμpq

  • Hall coefficient depends on the hole and electron concentration, and also on the mobility of carriers.
  • In an insulator, the gap between the conduction band and the valence band is very high.
  • Conductivity will be almost zero in the insulator
  • As conductivity zero in insulator then mobility also almost zero.
  • So, the hall coefficient will be zero almost for the insulator.
Test: Hall Effect - Question 6

A hall’s effect pick-up can be used for measuring ________. 

Detailed Solution for Test: Hall Effect - Question 6

Hall Effect Transducers:

  • when a conductor is kept perpendicular to the magnetic field and a direct current is passed through it, It results in an electric field perpendicular to the directions of both the magnetic field and current with a magnitude proportional to the product of the magnetic field strength and current. The voltage so developed is very small and it is difficult to detect it. But in semiconductors such as germanium, this voltage is enough for measurement with a sensitive moving coil instrument. This phenomenon is called the Hall effect.
  • Commercial Hall effect transducers are made from germanium or other semiconductor materials.
  • Hall effect transducers find applications in instruments that measure magnetic field with small flux densities.
  • Hall effect pick up can be used for measurement of current by magnetic field produced due to flow of current.   
  • Hall effect element may be used for measuring a linear displacement or location of the structural elements in cases where it is possible to change the magnetic field strength by variation in geometry of the magnetic structure.
  • The main advantage of hall effect transducers is that they are non-contact device with small size and high resolution.
  • The main drawback of hall effect transducers is highly sensitive to temperature variation and variation of Hall coefficient from plate to plate, thereby requiring individual calibration in each case.
Test: Hall Effect - Question 7

What is the conductivity when the Hall Effect coefficient is 5 and mobility is 5cm2 /s.

Detailed Solution for Test: Hall Effect - Question 7

µ=σR
σ =µ/R
=5*10-4/5
=0.0001 S/m.

Test: Hall Effect - Question 8

The Hall voltage (Vₕ) is given by the formula:

Detailed Solution for Test: Hall Effect - Question 8

The Hall voltage is determined using the following formula:

  • B is the magnetic field strength.
  • I is the current flowing through the conductor.
  • n is the charge carrier density.
  • e is the elementary charge (charge of an electron).
  • w is the width of the conductor.

Formula for Hall voltage (VH) is:

VH = (B × I) / (n × e × w)

Test: Hall Effect - Question 9

The Hall coefficient (Rₕ) for an intrinsic semiconductor is:

Detailed Solution for Test: Hall Effect - Question 9

The Hall coefficient for an intrinsic semiconductor is zero. This is because:

  • Intrinsic semiconductors have an equal number of electrons and holes.
  • The contributions of electrons and holes to the Hall effect cancel each other out.
  • This balance results in no net Hall voltage.
Test: Hall Effect - Question 10

If the current in a Hall effect experiment is doubled while keeping other parameters constant, the Hall voltage will:

Detailed Solution for Test: Hall Effect - Question 10

In a Hall effect experiment:

  • The Hall voltage is directly proportional to the current.
  • If the current is doubled, the Hall voltage will also double.

Therefore, the correct answer is that the Hall voltage will double.

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