AP Physics 2- 12.3 Magnetism and Current- Carrying Wires- Study Notes- New Syllabus
AP Physics 2- 12.3 Magnetism and Current- Carrying Wires – Study Notes
AP Physics 2- 12.3 Magnetism and Current- Carrying Wires – Study Notes – per latest Syllabus.
Key Concepts:
- Magnetic Field Produced by a Current-Carrying Wire
- Magnetic Force on a Current-Carrying Wire in a Magnetic Field
Magnetic Field Produced by a Current-Carrying Wire
A moving charge or current produces a magnetic field in the surrounding space. The direction of the magnetic field around a straight current-carrying wire is determined by the right-hand rule: if the thumb points in the direction of current, the curled fingers show the circular magnetic field lines.
Biot–Savart Law (General Expression):
The magnetic field at a point due to a small current element is given by:
\( \delta \vec{B} = \dfrac{\mu_0}{4\pi} \dfrac{I \, \delta \vec{l} \times \hat{r}}{r^2} \)
- \(\mu_0\): permeability of free space \((4\pi \times 10^{-7} \, \mathrm{T \cdot m/A})\).
- \(I\): current in the wire.
- \(\delta \vec{l}\): small element of the wire in the direction of current.
- \(\hat{r}\): unit vector pointing from the current element to the field point.
- \(r\): distance between the current element and the field point.
Magnetic Field Due to a Long Straight Wire:
Using the Biot–Savart law and symmetry, the magnetic field at a perpendicular distance \(\mathrm{r}\) from a long straight wire is:
\( B = \dfrac{\mu_0 I}{2\pi r} \)
- Direction is given by the right-hand rule.
- Field lines form concentric circles around the wire.
Magnetic Field of a Circular Current Loop:
At the center of a circular loop of radius \(\mathrm{R}\) carrying current \(\mathrm{I}\):
\( B = \dfrac{\mu_0 I}{2R} \)
- Field direction: given by the right-hand rule (curl fingers along current → thumb gives field direction at center).
- For \(N\) turns: \( B = \dfrac{\mu_0 N I}{2R} \).
Magnetic Field Inside a Long Solenoid:
A solenoid is a coil of many turns of wire carrying current. The magnetic field inside an ideal solenoid is nearly uniform:
\( B = \mu_0 n I \)
- \(n = \dfrac{N}{L}\): number of turns per unit length.
- Outside the solenoid, the field is very weak (ideal solenoid approximation).
Example :
A long straight wire carries a current of \(5 \, \mathrm{A}\). Find the magnetic field at a distance of \(2 \, \mathrm{cm}\) from the wire.
▶️ Answer/Explanation
Step 1: Formula: \( B = \dfrac{\mu_0 I}{2\pi r} \).
Step 2: Substituting values: \( B = \dfrac{(4\pi \times 10^{-7})(5)}{2\pi (0.02)} \).
Step 3: Simplify: \( B = \dfrac{20\pi \times 10^{-7}}{0.04\pi} = 5 \times 10^{-5} \, \mathrm{T} \).
Answer: \( B = 5.0 \times 10^{-5} \, \mathrm{T} \), directed in circular loops around the wire.
Force on a Current-Carrying Wire in a Magnetic Field
A moving charge in a magnetic field experiences a force given by the Lorentz force:
\(\mathrm{ \vec{F} = q \, \vec{v} \times \vec{B} }\)
Since an electric current consists of many moving charges, a wire carrying current in a magnetic field also experiences a force.
Force on a Straight Current-Carrying Wire:
For a straight wire of length \(\mathrm{L}\), carrying current \(\mathrm{I}\) in a uniform magnetic field \(\vec{B}\):
\( \vec{F} = I \, \vec{L} \times \vec{B} \)
- \(\mathrm{I}\): current in the wire.
- \(\vec{L}\): vector in the direction of current, with magnitude equal to wire length.
- \(\vec{B}\): magnetic field vector.
Magnitude of the Force:
\( F = I L B \sin \theta \)
- \(\theta\): angle between wire (current direction) and magnetic field.
- Maximum force when wire is perpendicular to \(\vec{B}\).
- No force if wire is parallel to \(\vec{B}\).
Direction of the Force:
- Given by the Right-Hand Rule for Force:
- Point fingers in direction of current (\(\vec{I}\)).
- Point palm toward direction of \(\vec{B}\).
- Thumb gives direction of force on the wire.
Applications:
- Electric motors: Force on current loops causes rotational motion.
- Rail guns: A conducting rod placed across rails carrying current experiences force, propelling it forward.
Example :
A \(0.25 \, \mathrm{m}\) long wire carries a current of \(3.0 \, \mathrm{A}\) perpendicular to a uniform magnetic field of \(0.40 \, \mathrm{T}\). Find the magnitude of the force on the wire.
▶️ Answer/Explanation
Step 1: Formula: \( F = I L B \sin \theta \).
Step 2: Substituting values: \( F = (3.0)(0.25)(0.40)\sin 90^\circ \).
Step 3: \( F = 0.30 \, \mathrm{N} \).
Answer: The force on the wire is \( 0.30 \, \mathrm{N} \), direction given by right-hand rule.