AP Physics C E&M- 12.2 Magnetism and Moving Charges- Study Notes- New Syllabus
AP Physics C E&M- 12.2 Magnetism and Moving Charges – Study Notes
AP Physics C E&M- 12.2 Magnetism and Moving Charges – Study Notes – per latest Syllabus.
Key Concepts:
- Magnetic Field Produced by Moving Charged Objects
- Force on Moving Charges in a Magnetic Field
- Motion of Charges in Electric and Magnetic Fields
- The Hall Effect
Magnetic Field Produced by a Moving Charged Object
A single moving charged particle produces a magnetic field in the surrounding space. This field depends on the particle’s charge, velocity, and the position of the observation point relative to the particle.
Key Properties:
- The magnetic field depends on the velocity of the charge and the distance from the charge to the observation point.
- The direction of the field at any point is:
- Perpendicular to the particle’s velocity vector \(\vec{v}\).
- Perpendicular to the position vector \(\vec{r}\) from the particle to the observation point.
- Given by the right-hand rule for \(\vec{v} \times \vec{r}\).
- The magnetic field is maximum when \(\vec{v}\) and \(\vec{r}\) are perpendicular, and zero when they are parallel.
Biot–Savart Law for a Moving Point Charge:
The magnetic field at a point in space due to a point charge \(q\) moving with velocity \(\vec{v}\) is:
\( \mathrm{\vec{B} = \dfrac{\mu_0}{4 \pi} \dfrac{q \, (\vec{v} \times \hat{r})}{r^2}} \)
- \(\mathrm{q}\) = charge of the particle
- \(\mathrm{\vec{v}}\) = velocity of the charge
- \(\mathrm{r}\) = distance from the charge to the observation point
- \(\mathrm{\hat{r}}\) = unit vector from the charge to the point
Key Idea: A moving charge produces a magnetic field that encircles its path. The field is strongest when the motion is perpendicular to the line connecting the charge and the observation point.
Example :
Find the direction of the magnetic field at a point located to the east of a proton moving upward (north to south is ignored).
▶️ Answer/Explanation
Step 1: Proton is moving upward (\(\vec{v}\)).
Step 2: The position vector from proton to point is toward the east (\(\vec{r}\)).
Step 3: \(\vec{v} \times \vec{r}\) points into the page (right-hand rule).
Final Answer: The magnetic field points into the page.
Example :
A particle with charge \( q = 1.6 \times 10^{-19} \, C \) moves at \( v = 3.0 \times 10^6 \, m/s \). Find the magnetic field at a point \( 0.05 \, m \) away, located such that \(\vec{v}\) and \(\vec{r}\) are perpendicular.
▶️ Answer/Explanation
Step 1: Formula: \( \mathrm{B = \dfrac{\mu_0}{4 \pi} \dfrac{q v \sin \theta}{r^2}} \).
Step 2: Since \(\theta = 90^\circ\), \(\sin \theta = 1\).
Step 3: Substitute: \( \mathrm{B = (1.0 \times 10^{-7}) \dfrac{(1.6 \times 10^{-19})(3.0 \times 10^6)}{(0.05)^2}} \).
Step 4: \( \mathrm{B = 1.9 \times 10^{-17} \, T} \).
Final Answer: The magnetic field at the point is \( \mathrm{1.9 \times 10^{-17} \, T} \).
Force on Moving Charges in a Magnetic Field
A charged particle moving in a magnetic field experiences a force due to its motion relative to the field. This is called the magnetic force.
Mathematical Expression:
The force on a particle of charge \(q\) moving with velocity \(\mathrm{\vec{v}}\) in a magnetic field \(\mathrm{\vec{B}}\) is:
\( \mathrm{\vec{F_B} = q \, \vec{v} \times \vec{B}} \)
- Magnitude:
\( \mathrm{F_B = q v B \sin \theta} \)
where \(\theta\) is the angle between \(\vec{v}\) and \(\vec{B}\).
- Direction: given by the right-hand rule (for positive charges). For negative charges, the force is in the opposite direction.
Properties of the Magnetic Force:
- The force is always perpendicular to both velocity \(\vec{v}\) and magnetic field \(\vec{B}\).
- The magnetic force does no work because it is always perpendicular to displacement.
- The speed of the particle remains constant; only its direction changes.
Motion of Charges in Magnetic Fields:
- Perpendicular velocity (\(\theta = 90^\circ\)): The particle moves in a circle of radius
\( \mathrm{r = \dfrac{m v}{|q| B}} \)
- Parallel velocity (\(\theta = 0^\circ\)): No magnetic force acts (\(F_B = 0\)).
- General velocity: The particle follows a helical path (circular motion in the plane perpendicular to \(\vec{B}\) plus linear motion along \(\vec{B}\)).
Key Consequences:
- Magnetic fields can change the direction of a charge’s motion but not its speed.
- This principle is applied in cyclotrons, mass spectrometers, and particle accelerators.
Example :
An electron (\( q = -1.6 \times 10^{-19} \, C \)) enters a uniform magnetic field of \( 0.20 \, T \) with a speed of \( 3.0 \times 10^6 \, m/s \), moving perpendicular to the field. Find the radius of its circular path.
▶️ Answer/Explanation
Step 1: Radius formula: \( \mathrm{r = \dfrac{m v}{|q| B}} \)
Step 2: Substitute values (\( m_e = 9.11 \times 10^{-31} \, kg \)): \( \mathrm{r = \dfrac{(9.11 \times 10^{-31})(3.0 \times 10^6)}{(1.6 \times 10^{-19})(0.20)}} \)
Step 3: \( \mathrm{r \approx 8.5 \times 10^{-5} \, m = 0.085 \, mm} \)
Final Answer: The electron moves in a circle of radius \( \mathrm{8.5 \times 10^{-5} \, m} \).
Example :
A proton enters a magnetic field of \( 0.10 \, T \) at an angle of \( 30^\circ \) with a velocity of \( 2.0 \times 10^6 \, m/s \). Describe its motion.
▶️ Answer/Explanation
Step 1: Break velocity into components: \( \mathrm{v_\perp = v \sin 30^\circ = 1.0 \times 10^6 \, m/s} \), \( \mathrm{v_\parallel = v \cos 30^\circ \approx 1.73 \times 10^6 \, m/s} \).
Step 2: Perpendicular component causes circular motion with radius: \( \mathrm{r = \dfrac{m v_\perp}{q B}} \).
Step 3: Parallel component causes uniform motion along the field.
Final Answer: The proton follows a helical path, with circular motion of radius determined by \(\mathrm{v_\perp}\), while moving forward along \(\vec{B}\) with speed \(\mathrm{v_\parallel}\).
Motion of Charges in Electric and Magnetic Fields
Forces on a Moving Charge:
A moving charge in a region with both electric field \(\mathrm{\vec{E}}\) and magnetic field \(\mathrm{\vec{B}}\) experiences independent forces from each field.
Electric Force:
\( \mathrm{\vec{F_E} = q \vec{E}} \)
Magnetic Force:
\( \mathrm{\vec{F_B} = q \, \vec{v} \times \vec{B}} \)
Total Force (Lorentz Force Law):
\( \mathrm{\vec{F} = q (\vec{E} + \vec{v} \times \vec{B})} \)
Properties:
- Electric force acts in the direction of the electric field (for positive charges).
- Magnetic force acts perpendicular to both velocity and magnetic field.
- If \(\vec{E}\) and \(\vec{B}\) are perpendicular, a velocity exists for which the two forces cancel:
\( \mathrm{v = \dfrac{E}{B}} \)
This principle is used in velocity selectors.
The Hall Effect
The Hall effect is the creation of a transverse electric potential difference (Hall voltage) across a current-carrying conductor placed in a magnetic field perpendicular to the direction of current flow.
Cause:
- Charges moving in the conductor experience a magnetic force (\( q \vec{v} \times \vec{B} \)) that pushes them to one side of the conductor.
- This accumulation of charges creates an electric field, which produces a balancing electric force.
Hall Voltage:
\( \mathrm{V_H = \dfrac{IB}{n q t}} \)
- \(\mathrm{I}\) = current
- \(\mathrm{B}\) = magnetic field strength
- \(\mathrm{n}\) = number density of charge carriers
- \(\mathrm{q}\) = charge of each carrier
- \(\mathrm{t}\) = thickness of the conductor
Key Uses of the Hall Effect:
- Measuring magnetic field strength.
- Determining charge carrier density in a material.
- Identifying whether current is carried by positive or negative charges (sign of Hall voltage).
Example :
An electron (\(q = -1.6 \times 10^{-19} \, C\)) moves with velocity \( 2.0 \times 10^6 \, m/s \) through a region with electric field \( \vec{E} = 200 \, N/C \) (to the right) and magnetic field \( \vec{B} = 1.0 \times 10^{-4} \, T \) (into the page). Find the net force on the electron if its velocity is upward.
▶️ Answer/Explanation
Step 1 (Electric force): \( \mathrm{F_E = qE = (-1.6 \times 10^{-19})(200)} = -3.2 \times 10^{-17} \, N \) (leftward).
Step 2 (Magnetic force): \( \mathrm{F_B = q v B} = (1.6 \times 10^{-19})(2.0 \times 10^6)(1.0 \times 10^{-4}) \). \( \mathrm{F_B = 3.2 \times 10^{-17} \, N} \), direction by RHR = rightward. Since charge is negative, force is opposite = leftward.
Step 3 (Total force): Both forces are leftward, so \( \mathrm{F_{net} = 6.4 \times 10^{-17} \, N \, (left)} \).
Final Answer: Net force on the electron is \( \mathrm{6.4 \times 10^{-17} \, N} \) to the left.
Example :
A copper strip of thickness \(1.0 \, mm\) carries a current of \(3.0 \, A\). It is placed in a magnetic field of \(0.20 \, T\) perpendicular to the strip. The number density of conduction electrons in copper is \(8.5 \times 10^{28} \, m^{-3}\). Calculate the Hall voltage.
▶️ Answer/Explanation
Step 1: Formula: \( \mathrm{V_H = \dfrac{IB}{n q t}} \).
Step 2: Substitute: \( \mathrm{V_H = \dfrac{(3.0)(0.20)}{(8.5 \times 10^{28})(1.6 \times 10^{-19})(1.0 \times 10^{-3})}} \).
Step 3: \( \mathrm{V_H = 4.4 \times 10^{-9} \, V} \).
Final Answer: The Hall voltage is \( \mathrm{4.4 \, nV} \).