Problem 13 - Fundamentals of Physics

Question

A strip of copper \(150 \mu \mathrm{m}\) thick and \(4.5 \mathrm{~mm}\) wide is placed in a uniform magnetic field \(\vec{B}\) of magnitude \(0.65 \mathrm{~T},\) with \(\vec{B}\) perpendicular to the strip. A current \(i=23 \mathrm{~A}\) 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 \(\left./ \mathrm{m}^{3} .\right)\)

Step-by-Step Solution

Answer

The Hall voltage \( V \) is approximately \( 1.17 \mathrm{~\mu V} \).

Step 1: Understand the Hall Effect

The Hall effect occurs when a current flows through a conductor in the presence of a perpendicular magnetic field. A transverse voltage, called the Hall voltage, is generated across the conductor due to the force on the moving charges by the magnetic field.

Step 2: Determine the Known Values

Given parameters are:- Thickness of the copper strip, \( t = 150 \mu \mathrm{m} = 150 \times 10^{-6} \mathrm{~m} \)- Width of the strip, \( w = 4.5 \mathrm{~mm} = 4.5 \times 10^{-3} \mathrm{~m} \)- Magnetic field strength, \( B = 0.65 \mathrm{~T} \)- Current through the strip, \( i = 23 \mathrm{~A} \)- Charge carrier density, \( n = 8.47 \times 10^{28} \mathrm{~electrons/m}^3 \).

Step 3: Hall Voltage Formula

The Hall voltage \( V \) is given by:\[ V = \frac{Bi}{net} \]where:- \( B \) is the magnetic field strength,- \( i \) is the current,- \( n \) is the charge carrier density,- \( e \) is the elementary charge \((1.6 \times 10^{-19} \mathrm{~C})\),- \( t \) is the thickness of the conductor.

Step 4: Substitute and Calculate

Substitute the known values into the Hall voltage formula:\[V = \frac{(0.65 \mathrm{~T})(23 \mathrm{~A})}{(8.47 \times 10^{28} \mathrm{~electrons/m}^3)(1.6 \times 10^{-19} \mathrm{~C})(150 \times 10^{-6} \mathrm{~m})}\]Calculate this to find the Hall voltage \( V \).

Step 5: Perform the Calculation

Proceeding with the calculation:\[V = \frac{(0.65)(23)}{(8.47 \times 10^{28})(1.6 \times 10^{-19})(150 \times 10^{-6})}\]Simplify the calculation:\[ V \approx 1.17 \times 10^{-6} \mathrm{~V} = 1.17 \mathrm{~\mu V} \]

Key Concepts

Magnetic Field

A magnetic field represents a region where a magnetic force is observed. This force affects charged particles, like electrons, when they move within it.
When discussing the Hall effect, the magnetic field is the key component that causes charge carriers to deflect. This deflection results in a build-up of voltage across the conductor.
In our exercise, the magnetic field has a magnitude of 0.65 Tesla, positioned perpendicular to the copper strip.

• Magnetic Field Direction: It must be perpendicular to the current for the Hall effect to occur efficiently.
• Magnitude: The larger the magnetic field, the more noticeable the deflection of charge carriers will be.

Current

Current is the flow of electrical charge carriers, like electrons, through a conductor.
In the Hall effect, when current flows through a conductor in a magnetic field, it experiences a perpendicular force.
This force causes the charge carriers to accumulate on one side, creating a voltage across the conductor known as the Hall voltage.
For our copper strip, the current flowing is 23 amperes, moving straight across the strip’s length.

• Role in Hall Effect: The amount of current influences the magnitude of the Hall voltage.
• Direction: Current direction is key to determining which side of the strip the charge builds up on.

Charge Carrier Density

Charge carrier density refers to the number of charge carriers (usually electrons) per unit volume within a material.
In the case of our copper strip, the charge carrier density is given as \(8.47 \times 10^{28}\) electrons per cubic meter.
This value is crucial for calculating the Hall voltage because it indicates how many electrons are available to carry the charge through the material.

• Influence on Hall Voltage: Higher charge carrier densities can reduce the Hall voltage magnitude because more carriers are available to distribute the force from the magnetic field.
• Material Property: Different materials will have different carrier densities, affecting their suitability for Hall effect applications.

Hall Voltage

Hall voltage is the voltage generated across a conductor due to the Hall effect.
This occurs as charge carriers are deflected by a magnetic field while moving through a current.
The Hall voltage provides a measure of the magnetic field and can be calculated using the formula \( V = \frac{Bi}{net} \).
In this formula, \( B \) is the magnetic field strength, \( i \) is the current, \( n \) is the charge carrier density, \( e \) is the elementary charge, and \( t \) is the thickness of the conductor.

• Calculation: Substituting known values gives \( V \approx 1.17 \ \mu V \).
• Applications: Hall voltage is useful for measuring magnetic fields and aiding in the design of electronic sensors.