AP Physics C E&M- 12.1 Magnetic Fields- Study Notes- New Syllabus
AP Physics C E&M- 12.1 Magnetic Fields – Study Notes
AP Physics C E&M- 12.1 Magnetic Fields – Study Notes – per latest Syllabus.
Key Concepts:
- Properties of a Magnetic Field
- Magnetic Behavior of Materials
- Magnetic Permeability of a Material
Properties of a Magnetic Field
A magnetic field is a region of space where moving charges or magnetic dipoles experience a force. It is a vector field that describes the influence of magnetic forces.
Vector Nature:
- The magnetic field is represented by the vector \(\mathrm{\vec{B}}\).
- Its direction is defined as the direction of the force on the north pole of a magnetic compass.
- The magnitude is measured in teslas (\(\mathrm{T}\)), where:
\( \mathrm{1 \, T = 1 \, N \cdot s / (C \cdot m)} \)
Magnetic Field Lines:
- Magnetic field lines form closed loops: they emerge from the north pole and enter the south pole of a magnet, continuing through the magnet back to the north pole.
- Field lines are continuous and have no start or end point (unlike electric field lines).
- The density of lines represents the field’s strength.
- Field lines never intersect.
Gauss’s Law for Magnetism:
Gauss’s law for magnetism states that the net magnetic flux through any closed surface is zero:
\( \mathrm{\oint \vec{B} \cdot d\vec{A} = 0} \)
- This implies that magnetic monopoles do not exist; magnetic poles always come in north–south pairs.
- It also explains why magnetic field lines must always form closed loops.
Magnetic Dipoles:
- Current loops and bar magnets behave as magnetic dipoles with a magnetic moment \(\mathrm{\vec{\mu}}\).
- A dipole in a magnetic field experiences torque:
\( \mathrm{\vec{\tau} = \vec{\mu} \times \vec{B}} \)
- The dipole tends to align with the external magnetic field.
Sources of Magnetic Fields:
- Moving charges (currents).
- Permanent magnets (due to atomic magnetic moments).
- Changing electric fields (per Maxwell’s equations).
Important Properties:
- Magnetic fields exert forces only on moving charges or magnetic dipoles.
- No force is exerted on stationary charges.
- Magnetic monopoles have not been observed; all magnetic sources are dipolar.
Example
A proton (\( q = 1.6 \times 10^{-19} \, C \)) moves at \( 2.0 \times 10^6 \, m/s \) perpendicular to a uniform magnetic field of \( 0.50 \, T \). Find the magnetic force on the proton.
▶️ Answer/Explanation
Step 1: Force formula: \( \mathrm{F = q v B \sin \theta} \), with \(\theta = 90^\circ\).
Step 2: Substitute values: \( \mathrm{F = (1.6 \times 10^{-19})(2.0 \times 10^6)(0.50)} \)
Step 3: \( \mathrm{F = 1.6 \times 10^{-13} \, N} \)
Final Answer: The proton experiences a magnetic force of \( \mathrm{1.6 \times 10^{-13} \, N} \).
Example
An electron moves parallel to a uniform magnetic field of \( 0.20 \, T \) at a speed of \( 1.0 \times 10^7 \, m/s \). What magnetic force acts on the electron?
▶️ Answer/Explanation
Step 1: Force formula: \( \mathrm{F = q v B \sin \theta} \).
Step 2: Since velocity is parallel to the field, \(\theta = 0^\circ\), so \(\sin 0 = 0\).
Step 3: \( \mathrm{F = 0} \).
Final Answer: No magnetic force acts on the electron.
Magnetic Behavior of Materials
The magnetic behavior of a material depends on how the magnetic dipoles (tiny atomic current loops due to electron motion and spin) are arranged and aligned inside the material.
Magnetic Dipoles:
- Each atom behaves like a small magnetic dipole due to the orbital motion and spin of its electrons.
- The magnetic moment of a dipole is represented by \(\mathrm{\vec{\mu}}\).
- In bulk materials, the alignment (or lack of alignment) of these dipoles determines the overall magnetic properties.
Types of Magnetic Materials:
Diamagnetic Materials:
- Dipoles are weakly induced in a direction opposite to the external magnetic field.
- Result: weak repulsion from magnetic fields.
- Examples: copper, bismuth, water.
- Magnetic susceptibility: \(\mathrm{\chi_m < 0}\).
Paramagnetic Materials:
- Have permanent dipoles, but they are randomly oriented without an external field.
- In the presence of an external field, dipoles align slightly with the field.
- Result: weak attraction to magnetic fields.
- Examples: aluminum, platinum, oxygen.
- Magnetic susceptibility: \(\mathrm{\chi_m > 0}\), but small.
Ferromagnetic Materials:
- Strong permanent dipoles spontaneously align in regions called domains.
- In an external field, domains grow and align, leading to very strong magnetization.
- Result: strong attraction to magnetic fields; can retain magnetization (hysteresis).
- Examples: iron, cobalt, nickel.
- Magnetic susceptibility: very large positive value.
Macroscopic Properties:
- Magnetization vector:
\( \mathrm{\vec{M} = \dfrac{\sum \vec{\mu}}{V}} \)
where \(\mathrm{V}\) is the volume of the material.
- Relation between \(\mathrm{\vec{B}}\), \(\mathrm{\vec{H}}\), and \(\mathrm{\vec{M}}\):
\( \mathrm{\vec{B} = \mu_0 (\vec{H} + \vec{M})} \)
- Effective permeability:
\( \mathrm{\mu = \mu_0 (1 + \chi_m)} \)
where \(\mathrm{\chi_m}\) is the magnetic susceptibility of the material.
Key Idea: The bulk magnetic properties of a material (diamagnetic, paramagnetic, or ferromagnetic) are determined by how atomic dipoles align under the influence of an external magnetic field.
Example :
Explain why iron is strongly attracted to a magnet, while copper is not.
▶️ Answer/Explanation
Step 1: Copper is diamagnetic: dipoles are induced opposite to the external field, producing a very weak repulsion.
Step 2: Iron is ferromagnetic: domains of dipoles align strongly with the field, producing a strong net magnetization.
Final Answer: Iron is strongly attracted to magnets because of domain alignment; copper shows almost no magnetic effect.
Example :
Which material will retain magnetization after the external magnetic field is removed: aluminum, bismuth, or cobalt?
▶️ Answer/Explanation
Step 1: Bismuth is diamagnetic → no permanent dipoles.
Step 2: Aluminum is paramagnetic → weak dipole alignment only in the presence of a field, does not retain magnetization.
Step 3: Cobalt is ferromagnetic → domains align and remain aligned even after the field is removed.
Final Answer: Cobalt retains magnetization (permanent magnet behavior).
Magnetic Permeability of a Material
Magnetic permeability describes how a material responds to an applied magnetic field. It measures the ability of a material to support the formation of a magnetic field within itself.
Key Relation:
The magnetic flux density \(\mathrm{\vec{B}}\) in a material is related to the magnetic field strength \(\mathrm{\vec{H}}\) by:
\( \mathrm{\vec{B} = \mu \vec{H}} \)
- \(\mathrm{\vec{B}}\) = magnetic flux density (tesla, T)
- \(\mathrm{\vec{H}}\) = magnetic field intensity (A/m)
- \(\mathrm{\mu}\) = permeability of the material (H/m)
Absolute and Relative Permeability:
- Absolute permeability: \(\mathrm{\mu}\) is the actual permeability of the material.
- Relative permeability: Ratio of permeability of a material to the permeability of free space:
\( \mathrm{\mu_r = \dfrac{\mu}{\mu_0}} \)
where \(\mathrm{\mu_0 = 4\pi \times 10^{-7} \, H/m}\) is the permeability of free space.
Connection with Susceptibility:
Permeability is related to magnetic susceptibility \(\mathrm{\chi_m}\):
\( \mathrm{\mu = \mu_0 (1 + \chi_m)} \)
- \(\mathrm{\chi_m < 0}\) → diamagnetic materials (\(\mu_r < 1\))
- \(\mathrm{\chi_m > 0}\) → paramagnetic materials (\(\mu_r > 1\), slightly)
- \(\mathrm{\chi_m \gg 1}\) → ferromagnetic materials (very large \(\mu_r\))
Properties of Magnetic Permeability:
- Determines how easily magnetic field lines pass through a material.
- Vacuum and air have \(\mathrm{\mu_r \approx 1}\).
- Ferromagnetic materials (iron, cobalt, nickel) have very high permeability → strong magnetization.
- In nonlinear materials, \(\mu\) is not constant and depends on field strength (hysteresis behavior).
Example :
A material has a relative permeability of \(\mu_r = 800\). If the applied magnetic field intensity is \( H = 200 \, A/m \), calculate the magnetic flux density \( B \).
▶️ Answer/Explanation
Step 1: Absolute permeability: \( \mathrm{\mu = \mu_r \mu_0 = 800 (4\pi \times 10^{-7}) = 1.005 \times 10^{-3} \, H/m} \)
Step 2: Magnetic flux density: \( \mathrm{B = \mu H = (1.005 \times 10^{-3})(200)} = 0.201 \, T \)
Final Answer: \( \mathrm{B \approx 0.20 \, T} \).
Example :
Air has \(\mu_r \approx 1\). If \(\chi_m = 2 \times 10^{-5}\) for aluminum, find the relative permeability of aluminum.
▶️ Answer/Explanation
Step 1: Relation: \( \mathrm{\mu_r = 1 + \chi_m} \)
Step 2: Substitute: \( \mathrm{\mu_r = 1 + 2 \times 10^{-5} = 1.00002} \)
Final Answer: Aluminum has relative permeability \(\mu_r \approx 1.00002\), very close to unity.