19.4 Current Density
19.4.1 Current Density
1. Definition
Figure 19.6 Variation of current through different small areas on a conductor cross-section. The amounts of charge Δq₁ and Δq₂ passing through equal areas ΔA in the same time interval Δt are different, resulting in different currents ΔI₁ and ΔI₂. This shows that current is not uniformly distributed across the cross-section and leads to the concept of current density (J).
Consider two small areas, each of value ΔA, that lie on the perpendicular cross-section of a conductor. A current, not necessarily constant, flows in the conductor. In general, the amount of charge Δq1 flowing in short time interval Δt perpendicularly through one small area ΔA on one part of a cross section may differ from the amount of charge Δq2 flowing in the same time interval simultaneously but on another part of the cross section, as shown in Figure 19.6. This means that the current ΔI1 (= Δq1/Δt) through one small area will differ from the current ΔI2 (= Δq2/Δt) through the other small area. In order to take into account such possible variation of Δq over the entire cross section of the conductor, we create a physical quantity known as the current density J at a point on the cross section.
It is defined as
$J=\lim_{\Delta A \to 0} \frac{\Delta I}{\Delta A}$
2. Unit
The unit of J is A m−2.
3. Quantity
Current density is a vector. It points in the direction along which a positive charge would move.
19.4.2 Current Density J for Constant Current I
Suppose the electric current I passing perpendicularly through a cross-sectional area A in a conducting substance is constant. In this case, we assume that charge Δq flowing through ΔA is the same on any part of the cross section, and the direction of charge flow at any point is perpendicular to the cross section. Then the magnitude of the current density J at any point on the cross section is the same and is given by
$\boxed{J=\frac{I}{A}}$
19.4.3 Relation between J and vd
1. Equation
$\boxed{v_d=\frac{I}{nqA}}$
Magnitude of current density:
$J=\frac{I}{A}=nqv_d$
We may write this expression in vector format, as follows:
$\boxed{J=nqv_d}$
2. Direction of J
(a) If the charge carriers are positively charged, then
$\boxed{J=(nq)v_d}$
J points in the direction of the positive charge flow.
(b) If the charge carriers are electrons, q will become q = −e. Then we have
$\boxed{J=-(ne)v_d}$
Now J points in a direction opposite to the direction of electron flow.
Example 19.4
Refer to Example 19.2. Determine the current density at a cross section of the copper wire.
Answer
I = 1.0 A
A = 7.85 × 10−7 m2
$J = \frac{1.0}{7.85 \times 10^{-7}}$ = 1.3 × 106 A m−2
The direction of J is perpendicular to the cross section. Since the uniform cross section is perpendicular to the length of the copper wire, then J points along the length of the wire.
Example 19.5
A uniform copper wire has a cross-sectional area of 2.0 mm2. A current of 3.0 A flows through it. Assuming that copper has 1 × 1029 free electrons per m3, determine:
(a) the drift velocity of the conduction electrons
(b) the current density
Answer
(a) $v_d=\frac{I}{nqA}=\frac{3.0}{(10^{29})(1.6 \times 10^{-19})(2.0 \times 10^{-6})}$
(b) $J=\frac{I}{A}=\frac{3.0}{2.0 \times 10^{-6}}=1.5 \times 10^6 \ A \ m^{-2}$
EXAMPLE 19.6
A straight electron beam of cross sectional area
1 × 10-8 m² strikes a flat screen perpendicularly.
The current density on the screen is
2 × 105 A m-2. Estimate:
(a) the current, assumed to be constant, produced by the electron beam
(b) the number of electrons striking the screen per second
(Electronic charge = −1.6 × 10-19 C)
Answer
(a) $J = \frac{I}{A}$
$I = J A$
$= (2 \times 10^5)(1 \times 10^{-8})$
$I= 2 \times 10^{-3} \ A$
(b) $I = \frac{\Delta Q}{\Delta t}$
$I= \frac{ne}{t} = \left(\frac{n}{\Delta t}\right)e$
$\left(\frac{n}{\Delta t}\right)=\frac{I}{e}$
$= \frac{2 \times 10^{-3}}{1.6 \times 10^{-19}} \cong 1 \times 10^{16} \ s^{-1}$
EXAMPLE 19.7
The drift velocity of electrons in a uniform wire of length 2.0 cm is 3.0 × 10-4 m s-1. If the number of free electrons in the 2.0 cm wire is 4.0 × 1021, determine the current in the wire. (Electronic charge = −1.6 × 10-19 C)
Jawab
Use the formula:
$I=nqv_d$
However, since total electrons are given:
$I = \frac{Q}{t}$
Step 1: Charge per electron
$e = 1.6 \times 10^{-19} \ C$
Step 2: Total charge
$Q = ne = (4.0 \times 10^{21})(1.6 \times 10^{-19}) = 6.4 \times 10^2 \ C$
Step 3: Time taken
$t = \frac{length}{velocity}= \frac{0.02}{3.0 × 10^{-4}} = 66.7 \ s$
Step 4: Current
$I = \frac{Q}{t} = \frac{6.4 \times 10^2}{66.7} \cong 9.6 \ A$
EXAMPLE 19.8
A constant current of 4.8 A flows in a uniform wire of cross sectional area 1.0 mm². About 1029 free electrons per m³ are available in the wire.
(a) Drift Velocity
Formula:
$I=nqv_d$
Rearrange:
$v_d=\frac{I}{nqA}$
Convert area:
A = 1.0 mm² = 1.0 × 10-6 m²
Substitute values:
$v_d= \frac{4.8}{(10^{29})(1.6 \times 10^{-19})(1.0 \times 10^{-6}}$
$vd = \frac{4.8}{1.6 \times 10^4} = 3.0 \times 10^{-4} \ m \ s^{-1}$
(b) Current Density
Formula:
$J = \frac{I}{A}$
$J = \frac{4.8}{1.0 \times 10^{-6}} = 4.8 \times 10^6 \ A \ m^{-2}$
EXAMPLE 19.9
The constant current and diameter of wire X and wire Y are given below.
Comparison of Wire X and Wire Y
| Property | Wire X | Wire Y |
|---|---|---|
| Current | I | I |
| Diameter | d | 2d |
| Current Density | JX | JY |
Note: This table compares current density in two wires with different diameters, a common concept in physics and electrical engineering.
Determine the ratio $J_X/JY$.
Answer
Current density is given by:
$J = \frac{I}{A}$
The cross-sectional area of a wire is:
$A = \pi \left(\frac{d}{2}\right)^2= \pi \frac{d^2}{4}$
For wire X:
$A_X= \pi \frac{d^2}{4}$
For wire Y (diameter = 2d):
$A_Y= \pi \frac{(2d)^2}{4}=\pi d^2$
Thus,
$J_X=\frac{I}{A_X}=\frac{4I}{\pi d^2}$
$J_Y=\frac{I}{A_Y}=\frac{I}{\pi d^2}$
Therefore, the ratio is:
$\frac{J_X}{J_Y}=4$
Applications of Current Density
Current density is an important concept in electricity and electronics, as it helps explain how electric current is distributed within a conductor. It has many practical applications in science, engineering, and technology.
1. Electrical Power Transmission
Current density is used to determine the safe current-carrying capacity of transmission lines. High current density can cause overheating, so engineers design conductors with appropriate cross-sectional areas to prevent energy loss and damage.
2. Electronic Circuit Design
In electronic circuits, current density helps engineers design components such as wires and circuit boards to ensure efficient and safe current flow without overheating or failure.
3. Semiconductor Devices
Current density plays a key role in semiconductor devices such as diodes and transistors. It helps in understanding how charge carriers move within the material and affects the performance of electronic components.
4. Heating Effects (Joule Heating)
The heating effect of electric current depends on current density. Higher current density leads to greater heat production, which is useful in devices like electric heaters but must be controlled in sensitive electronic systems.
5. Electrolysis and Electroplating
In electrochemical processes such as electrolysis and electroplating, current density determines the rate of chemical reactions and the quality of the deposited material.
6. Material Science and Conductivity Analysis
Current density is used to study the electrical properties of materials, including conductivity and resistivity, helping scientists develop better conductive and semiconductor materials.
7. Safety in Electrical Systems
Understanding current density is important for preventing electrical hazards such as overheating, short circuits, and electrical fires by ensuring that conductors operate within safe limits.
Conclusion: Current Density
In conclusion, current density is an important physical quantity used to describe how electric current is distributed over a cross-sectional area of a conductor. It provides a more detailed understanding of electric current by considering how charge flows at a specific point rather than across the entire conductor.
The concept of current density becomes essential when the flow of charge is not uniform, as different regions of a conductor may carry different amounts of current. By defining current density as the current per unit area, we can analyze electrical conduction more accurately in various materials and conditions.
Current density is a vector quantity, meaning it has both magnitude and direction, and it points in the direction of positive charge flow. Its SI unit is ampere per square meter (A m−2).
Overall, understanding current density is crucial in the study of electricity and electronics, especially in fields such as circuit analysis, semiconductor physics, and electrical engineering applications.
Frequently Asked Questions (FAQ) – Current Density
1. What is current density?
Current density is the amount of electric current flowing per unit cross-sectional area of a conductor. It provides a detailed description of how current is distributed at a specific point in a material.
2. What is the formula for current density?
The formula for current density is:
J = I / A
where J is current density, I is electric current, and A is the cross-sectional area.
3. What is the SI unit of current density?
The SI unit of current density is ampere per square meter (A m−2).
4. Is current density a scalar or vector quantity?
Current density is a vector quantity because it has both magnitude and direction. Its direction is the same as the direction of positive charge flow.
5. Why is current density important?
Current density is important because it helps us understand how electric current is distributed within a conductor, especially when the current is not uniform across the cross-section.
6. What factors affect current density?
Current density depends on:
- the electric current (I)
- the cross-sectional area (A)
7. How is current density used in real life?
Current density is used in electrical engineering, power transmission, semiconductor devices, electrolysis, and circuit design to ensure safe and efficient current flow.
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