AP Physics C E&M- 11.1 Electric Current- Study Notes- New Syllabus
AP Physics C E&M- 11.1 Electric Current – Study Notes
AP Physics C E&M- 11.1 Electric Current – Study Notes – per latest Syllabus.
Key Concepts:
- Electric Current
- Current Density
- Direction of Electric Current
Electric Current
Electric current is the rate at which charge passes through a cross-sectional area of a conductor. Current arises from the collective motion of charge carriers (usually electrons) under the influence of an electric potential difference.
Equation for Current:
\( \mathrm{I = \dfrac{dq}{dt}} \)
- \(\mathrm{I}\) = current (A)
- \(\mathrm{dq}\) = amount of charge passing through a cross section
- \(\mathrm{dt}\) = time interval
Drift Velocity Model:
Although individual charge carriers move randomly, in the presence of an electric field, they acquire a small net velocity called the drift velocity.
\( \mathrm{I = n q v_d A} \)
- \(\mathrm{n}\) = number density of charge carriers (\( \mathrm{m^{-3}} \))
- \(\mathrm{q}\) = charge of each carrier (C)
- \(\mathrm{v_d}\) = drift velocity (\( \mathrm{m/s} \))
- \(\mathrm{A}\) = cross-sectional area of the conductor (\( \mathrm{m^2} \))
The drift velocity is typically very small (on the order of mm/s), but since the number of charge carriers is very large, the current can be significant.
Driving Force for Current:
- Charge moves in response to an electric potential difference across the conductor.
- This potential difference is sometimes called the electromotive force (emf), represented by \(\mathcal{E}\).
- Even though the emf is not a true force, it represents the energy supplied per unit charge to move charges through the circuit.
Zero Current Condition:
- If the current in a wire is zero (\(I = 0\)), the net drift velocity of charge carriers is also zero.
- However, individual charge carriers still move randomly at very high speeds due to thermal energy; these random motions cancel out on average.
Example
A copper wire has a cross-sectional area of \( 1.0 \times 10^{-6} \, m^2 \). It carries a current of \( 3.0 \, A \). If the number density of conduction electrons in copper is \( 8.5 \times 10^{28} \, m^{-3} \), calculate the drift velocity of the electrons.
▶️ Answer/Explanation
Step 1: Use formula: \( \mathrm{I = n q v_d A} \).
Step 2: Solve for drift velocity: \( \mathrm{v_d = \dfrac{I}{n q A}} \).
Step 3: Substitute values: \( \mathrm{v_d = \dfrac{3.0}{(8.5 \times 10^{28})(1.6 \times 10^{-19})(1.0 \times 10^{-6})}} \).
Step 4: \( \mathrm{v_d \approx 2.2 \times 10^{-4} \, m/s} \).
Final Answer: The drift velocity is extremely small, about \( \mathrm{0.22 \, mm/s} \).
Example
A wire carries a constant current of \( I = 2.0 \, A \). How much charge passes through a cross section of the wire in \( 1.5 \, minutes \)?
▶️ Answer/Explanation
Step 1: Current definition: \( \mathrm{I = \dfrac{q}{t}} \).
Step 2: Solve for charge: \( \mathrm{q = I t} \).
Step 3: Convert time: \( 1.5 \, min = 90 \, s \).
Step 4: Substitute: \( \mathrm{q = (2.0)(90) = 180 \, C} \).
Final Answer: A total of \( \mathrm{180 \, C} \) of charge passes through the wire in 1.5 minutes.
Current Density
Current density is the amount of electric current flowing per unit cross-sectional area of a conductor. It is a vector quantity that points in the direction of the current flow (conventional current, positive charge flow).
Equation for Current Density:
\( \mathrm{\vec{J} = \dfrac{I}{A} \, \hat{n}} \)
- \(\mathrm{\vec{J}}\) = current density vector (\( \mathrm{A/m^2} \))
- \(\mathrm{I}\) = current (A)
- \(\mathrm{A}\) = cross-sectional area (\( \mathrm{m^2} \))
- \(\mathrm{\hat{n}}\) = unit vector in the direction of current
Relation to Drift Velocity:
From the drift velocity model, current density can also be written as:
\( \mathrm{\vec{J} = n q \vec{v_d}} \)
- \(\mathrm{n}\) = number density of charge carriers
- \(\mathrm{q}\) = charge of each carrier
- \(\mathrm{\vec{v_d}}\) = drift velocity of carriers
Relation to Electric Field:
When a potential difference is applied across a conductor, an electric field \(\mathrm{\vec{E}}\) is established inside it. Ohm’s microscopic law relates current density and the electric field:
\( \mathrm{\vec{J} = \sigma \vec{E}} \quad \text{or} \quad \vec{E} = \rho \vec{J} \)
- \(\mathrm{\sigma}\) = electrical conductivity of the material
- \(\mathrm{\rho = 1/\sigma}\) = resistivity of the material
Current from Current Density:
If the current density is non-uniform across the conductor, the total current can be determined by integrating over the cross-sectional area:
\( \mathrm{I = \displaystyle \int_A \vec{J} \cdot d\vec{A}} \)
- \(\mathrm{d\vec{A}}\) = infinitesimal area vector perpendicular to the surface
- The dot product ensures only the component of \(\vec{J}\) normal to the surface contributes to current flow.
Key Idea: Current density links microscopic motion of charges (drift velocity) to macroscopic current, and integrating \(\vec{J}\) across the conductor’s cross-section gives the total current.
Example
A copper wire of cross-sectional area \(1.0 \times 10^{-6} \, m^2\) carries a current of \(3.0 \, A\). Find the current density in the wire.
▶️ Answer/Explanation
Step 1: Formula: \( \mathrm{J = \dfrac{I}{A}} \).
Step 2: Substitute: \( \mathrm{J = \dfrac{3.0}{1.0 \times 10^{-6}}} \).
Step 3: \( \mathrm{J = 3.0 \times 10^6 \, A/m^2} \).
Final Answer: Current density is \( \mathrm{3.0 \times 10^6 \, A/m^2} \).
Example
The current density across a circular conductor of radius \(R = 2.0 \, mm\) varies with radial distance as \( J(r) = J_0 \left(1 – \dfrac{r}{R}\right) \), where \( J_0 = 5.0 \times 10^6 \, A/m^2 \). Find the total current through the wire.
▶️ Answer/Explanation
Step 1: Use integral: \( \mathrm{I = \displaystyle \int_A J(r) \, dA} \).
Step 2: In cylindrical coordinates: \( \mathrm{dA = 2\pi r \, dr} \).
Step 3: Substitute: \( \mathrm{I = \int_0^R J_0 \left(1 – \dfrac{r}{R}\right) (2\pi r \, dr)} \).
Step 4: Evaluate: \( \mathrm{I = 2\pi J_0 \left[\dfrac{r^2}{2} – \dfrac{r^3}{3R}\right]_0^R} \).
Step 5: Simplify: \( \mathrm{I = 2\pi J_0 \left(\dfrac{R^2}{2} – \dfrac{R^2}{3}\right)} = \dfrac{\pi}{3} J_0 R^2 \).
Step 6: Substitute numbers: \( \mathrm{I = \dfrac{\pi}{3} (5.0 \times 10^6)(2.0 \times 10^{-3})^2} \).
Step 7: \( \mathrm{I \approx 21 \, A} \).
Final Answer: The total current through the wire is about \( \mathrm{21 \, A} \).
Direction of Electric Current
Scalar Nature of Current:
Electric current is defined as the rate of flow of charge, \( \mathrm{I = \dfrac{dq}{dt}} \).
- It is a scalar quantity because it is defined by a single magnitude and a chosen direction of flow.
- Although current is associated with a direction of charge motion, it does not obey the laws of vector addition and does not have vector components.
Conventional Current Direction:
- The direction of conventional current is chosen to be the direction a positive charge would move in the circuit.
- This convention was established historically, before electrons were discovered.
Actual Charge Carriers:
- In most circuits, the mobile charge carriers are electrons, which are negatively charged.
- Electrons move in the direction opposite to the conventional current.
Key Idea: Current is treated as a scalar with a defined flow direction (conventional current = positive charges), even though in reality, electrons (negative charges) are moving in the opposite direction.
Example
In a copper wire carrying current to the right, in which direction do the electrons actually move?
▶️ Answer/Explanation
Step 1: Conventional current direction = right.
Step 2: Electrons are negative charge carriers, so they move opposite to conventional current.
Final Answer: The electrons drift to the left.
Example
A current of \( 2.0 \, A \) flows upward through a vertical wire. Which way are the electrons moving, and what is the rate of electron flow (number of electrons per second)?
▶️ Answer/Explanation
Step 1: Conventional current is upward, so electrons move downward.
Step 2: Number of electrons per second: \( \mathrm{n = \dfrac{I}{e}} = \dfrac{2.0}{1.6 \times 10^{-19}} \).
Step 3: \( \mathrm{n \approx 1.25 \times 10^{19} \, electrons/s} \).
Final Answer: Electrons move downward at a rate of \( \mathrm{1.25 \times 10^{19}} \, \text{per second}. \)