Home / AP® Exam / AP® Physics C: Electricity and Magnetism / AP Physics C E&M – 12.1 Magnetic Fields- Study Notes

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

AP Physics C E&M-Concise Summary Notes- All Topics

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.

Scroll to Top