AP Physics 2- 12.2 Magnetism and Moving Charges- Study Notes- New Syllabus
AP Physics 2- 12.2 Magnetism and Moving Charges – Study Notes
AP Physics 2- 12.2 Magnetism and Moving Charges – Study Notes – per latest Syllabus.
Key Concepts:
- Magnetic Field Produced by Moving Charges
- Magnetic Force on a Moving Charge
- Forces on a Moving Charge in Electric and Magnetic Fields & the Hall Effect
Magnetic Field Produced by Moving Charges
Electric charges that are stationary produce only an electric field. When charges are in motion, they also produce a magnetic field. The strength and direction of this magnetic field depend on the velocity of the charge and the position of the observation point.
Magnetic Field Due to a Moving Point Charge:
The magnetic field at a point in space due to a moving charge \(\mathrm{q}\) with velocity \(\vec{v}\) is given by:
\( \vec{B} = \dfrac{\mu_0}{4 \pi} \, \dfrac{q \, \vec{v} \times \hat{r}}{r^2} \)
- \(\mathrm{q}\): moving charge
- \(\vec{v}\): velocity of the charge
- \(\hat{r}\): unit vector from charge to point of observation
- \(\mathrm{r}\): distance between charge and observation point
- \(\mu_0\): permeability of free space (\(4\pi \times 10^{-7} \, \mathrm{T \cdot m / A}\))
Magnetic Field of a Current (Many Moving Charges):
- A steady current is a continuous flow of moving charges.
- Therefore, a current in a wire also produces a magnetic field.
Example :
A proton moves with speed \(2.0 \times 10^6 \, \mathrm{m/s}\) perpendicular to a point located \(0.05 \, \mathrm{m}\) away. Find the magnitude of the magnetic field produced at that point.
▶️ Answer/Explanation
Step 1: Formula: \(\mathrm{B = \dfrac{\mu_0}{4 \pi} \dfrac{q v \sin \theta}{r^2}}\).
Step 2: Substituting values: \(\mathrm{q = 1.6 \times 10^{-19} \, C,\; v = 2.0 \times 10^6 \, m/s,\; r = 0.05 \, m,\; \sin \theta = 1}\).
\(\mathrm{B = \dfrac{(10^{-7})(1.6 \times 10^{-19})(2.0 \times 10^6)}{(0.05)^2}}\).
Step 3: \(\mathrm{B = 1.28 \times 10^{-17} \, T}\).
Answer: The magnetic field is \(\mathrm{1.28 \times 10^{-17} \, T}\).
Magnetic Force on a Moving Charge
A charged particle moving in a magnetic field experiences a force called the magnetic force. This force depends on the charge, the velocity of the particle, the magnetic field strength, and the angle between velocity and field.
Mathematical Expression:
\( \vec{F} = q \, \vec{v} \times \vec{B} \)
- \(\mathrm{q}\): charge of the particle
- \(\vec{v}\): velocity of the particle
- \(\vec{B}\): magnetic field vector
- \(\vec{F}\): magnetic force
Magnitude of the Force:
\( F = q v B \sin \theta \)
- \(\mathrm{\theta}\): angle between \(\vec{v}\) and \(\vec{B}\).
- If \(\mathrm{\theta = 0^\circ \; or \; 180^\circ}\): the force is zero (motion parallel/antiparallel to field).
- If \(\mathrm{\theta = 90^\circ}\): the force is maximum (\(F = q v B\)).
Direction of the Force:
- Given by the right-hand rule: Point fingers along \(\vec{v}\), curl towards \(\vec{B}\), thumb points in direction of force for a positive charge.
- For a negative charge, the force direction is opposite.
Work Done by Magnetic Force:
- The magnetic force is always perpendicular to the velocity of the charge.
- Therefore, it does no work: \( W = \vec{F} \cdot \vec{d} = 0 \)
- The speed and kinetic energy of the particle remain constant; only the direction of motion changes.
Motion of Charged Particles in a Uniform Magnetic Field:
- If \(\vec{v} \perp \vec{B}\): The particle moves in a circular path with radius: \( r = \dfrac{m v}{q B} \)
- Angular frequency (cyclotron frequency): \( \omega = \dfrac{q B}{m} \)
- Period of revolution: \( T = \dfrac{2 \pi m}{q B} \)
- If \(\vec{v}\) has components both parallel and perpendicular to \(\vec{B}\): The path is a helix (circular motion around field lines + uniform motion along field).
Example :
An electron (\(\mathrm{q = -1.6 \times 10^{-19} \, C}\)) moves with speed \(\mathrm{3 \times 10^6 \, m/s}\) perpendicular to a magnetic field of strength \(\mathrm{2 \times 10^{-3} \, T}\). Find the radius of its circular path.
▶️ Answer/Explanation
Step 1: Formula: \(\mathrm{r = \dfrac{m v}{q B}}\).
Step 2: Mass of electron: \(\mathrm{m = 9.11 \times 10^{-31} \, kg}\).
Step 3: Substituting values: \(\mathrm{r = \dfrac{(9.11 \times 10^{-31})(3 \times 10^6)}{(1.6 \times 10^{-19})(2 \times 10^{-3})}}\).
Step 4: \(\mathrm{r \approx 8.5 \times 10^{-3} \, m = 8.5 \, mm}\).
Answer: The radius of the path is \(\mathrm{8.5 \, mm}\).
Forces on a Moving Charge in Electric and Magnetic Fields & the Hall Effect
Motion in Electric and Magnetic Fields:
A charged particle moving in a region with both an electric field \(\vec{E}\) and a magnetic field \(\vec{B}\) experiences two independent forces:
\( \vec{F} = q \vec{E} + q \vec{v} \times \vec{B} \)
- \(\mathrm{q}\): charge of the particle
- \(\vec{E}\): electric field vector
- \(\vec{v}\): velocity of the particle
- \(\vec{B}\): magnetic field vector
Key Points:
- The electric force \(\mathrm{q\vec{E}}\) acts parallel (or antiparallel) to the electric field.
- The magnetic force \(\mathrm{q \vec{v} \times \vec{B}}\) acts perpendicular to both \(\vec{v}\) and \(\vec{B}\).
- If both fields are present, the resultant motion depends on the balance between the electric and magnetic forces.
Hall Effect:
When a conductor carrying current is placed in a magnetic field that is perpendicular to the direction of motion of charges, a potential difference develops across the conductor. This is called the Hall effect.
- Cause: The magnetic force pushes moving charge carriers to one side of the conductor, creating charge separation.
- The resulting electric field (Hall field) builds up until the electric force balances the magnetic force.
Balance Condition:
\( q E_{\mathrm{H}} = q v B \)
Hall Electric Field:
\( E_{\mathrm{H}} = v B \)
Hall Voltage:
\( V_{\mathrm{H}} = E_{\mathrm{H}} d = v B d \)
- \(\mathrm{d}\): thickness of the conductor in the direction of charge separation.
Applications of the Hall Effect:
- Determining whether the majority charge carriers are positive (holes) or negative (electrons).
- Measuring magnetic field strength.
- Used in Hall sensors (automotive, current sensors, proximity detection).
Example :
A copper strip of thickness \(\mathrm{2 \, mm}\) carries a current with charge carriers moving at an average drift velocity \(\mathrm{v = 5 \times 10^{-4} \, m/s}\). It is placed in a magnetic field \(\mathrm{B = 0.8 \, T}\) perpendicular to the strip. Find the Hall voltage across its thickness.
▶️ Answer/Explanation
Step 1: Formula: \(\mathrm{V_H = v B d}\).
Step 2: Substituting values: \(\mathrm{V_H = (5 \times 10^{-4})(0.8)(2 \times 10^{-3})}\).
Step 3: \(\mathrm{V_H = 8 \times 10^{-7} \, V}\).
Answer: The Hall voltage is \(\mathrm{0.8 \, \mu V}\).