Unit 4 Magnetism: Understanding Magnetic Fields and the Forces They Produce

Magnetic Fields

What a magnetic field is (and what it is not)

A magnetic field is a vector field that exists in a region of space and can exert forces on moving electric charges and on currents. In AP Physics 2, the magnetic field is represented by the vector \vec{B} and measured in teslas (T).

It’s helpful to contrast magnetism with electrostatics:

  • Electric fields \vec{E} exert forces on charges whether the charges are moving or not.
  • Magnetic fields \vec{B} exert forces only on moving charges (or equivalently, on currents made of moving charges).

A common misconception is that “magnets pull on any charge.” They don’t. A stationary charge near a bar magnet experiences no magnetic force from the magnet’s field. Magnetism shows up when charges move.

Why magnetic fields matter

Magnetic fields are the “bridge” between electricity and motion. Once you understand \vec{B}, you can explain:

  • Why a compass needle turns near a wire with current
  • Why charged particles spiral in particle accelerators
  • How electric motors produce rotation
  • How velocity selectors and mass spectrometers separate particles

Magnetic fields also connect directly to later topics in Unit 4 (electromagnetic induction): changing magnetic fields are what generate induced electric fields and currents.

How we represent magnetic fields: field lines and direction conventions

We often draw magnetic field lines to visualize \vec{B}:

  • The direction of \vec{B} at a point is tangent to the field line.
  • Closer field lines mean stronger field.
  • Magnetic field lines form closed loops (they do not start or stop in empty space).

For bar magnets, field lines outside the magnet go from the north pole to the south pole, then return through the magnet, forming a loop.

Sources of magnetic fields

In AP Physics 2, the major sources you’re expected to know are:

  1. Permanent magnets (microscopic electric currents and magnetic domains)
  2. Electric currents in wires (moving charges)

A deep idea sits underneath: magnetism is fundamentally linked to moving charge.

Magnetic fields produced by currents (key geometries)

You typically won’t derive these in AP Physics 2, but you must know how to use them and interpret them.

Long straight wire

A long straight current-carrying wire produces magnetic field lines that form concentric circles around the wire.

Magnitude at distance r from the wire:

B = \frac{\mu_0 I}{2\pi r}

  • B is magnetic field magnitude (T)
  • \mu_0 is the permeability of free space
  • I is current (A)
  • r is perpendicular distance from the wire (m)

The constant:

\mu_0 = 4\pi \times 10^{-7}\ \text{T m/A}

Direction (right-hand rule): point your right thumb in the direction of conventional current; your curled fingers show the direction of \vec{B} circling the wire.

Common pitfall: reversing electron flow vs conventional current. In AP, right-hand rules use conventional current direction.

Single circular loop (field at the center)

At the center of a circular loop of radius R:

B = \frac{\mu_0 I}{2R}

For N identical loops stacked:

B = \frac{\mu_0 N I}{2R}

Direction: use a right-hand rule—curl fingers with current around the loop; thumb gives \vec{B} through the center.

Long solenoid (ideal)

A solenoid is a long coil of wire. Inside an ideal solenoid, the field is approximately uniform.

B = \mu_0 n I

  • n is turns per unit length (turns/m)

Direction: curl fingers in current direction around the coil; thumb points along \vec{B} inside.

Conceptual warning: students often assume the field outside a solenoid is equally strong. For an ideal long solenoid, the outside field is much weaker than the inside field.

Worked example: field near a straight wire

A long straight wire carries I = 5.0\ \text{A}. Find B at r = 2.0\ \text{cm}.

Step 1: Identify the model. Long straight wire, so use B = \mu_0 I / (2\pi r).

Step 2: Convert units.

r = 2.0\ \text{cm} = 2.0\times 10^{-2}\ \text{m}

Step 3: Compute.

B = \frac{(4\pi\times 10^{-7})(5.0)}{2\pi(2.0\times 10^{-2})}

Cancel \pi and simplify:

B = \frac{(20\times 10^{-7})}{(4.0\times 10^{-2})}

B = 5.0\times 10^{-5}\ \text{T}

So B = 5.0\times 10^{-5}\ \text{T} (about 50 microteslas).

Exam Focus

  • Typical question patterns:
    • Compute B from a wire, loop, or solenoid at a specific point, often with unit conversions.
    • Determine the direction of \vec{B} using a right-hand rule diagram.
    • Compare field strengths when I, r, N, or n changes.
  • Common mistakes:
    • Mixing up r (distance from wire) with diameter or radius given in the diagram.
    • Using electron flow direction instead of conventional current for right-hand rules.
    • Assuming field lines “start” at north and “end” at south without recognizing they form closed loops.

Magnetic Force on Moving Charges

What the magnetic force is

A magnetic force is the force a magnetic field exerts on a moving charged particle. The fundamental relationship is the magnetic part of the Lorentz force:

\vec{F} = q\vec{v} \times \vec{B}

  • \vec{F} is magnetic force (N)
  • q is charge (C)
  • \vec{v} is particle velocity (m/s)
  • \vec{B} is magnetic field (T)
  • \times indicates a cross product (direction matters)

The magnitude is:

F = |q| v B \sin\theta

where \theta is the angle between \vec{v} and \vec{B}.

Why this matters: “sideways” forces and circular motion

The most important qualitative idea is that magnetic force is perpendicular to both \vec{v} and \vec{B}. That means:

  • A magnetic force can change the direction of motion without changing speed (in many ideal cases).
  • Magnetic fields can cause circular or helical trajectories.

This is the physics behind particle beams bending in accelerators and charged particles being trapped in Earth’s magnetic field.

How it works: direction and right-hand rule

Because the force comes from a cross product, direction is not “toward the magnet.” Instead, you use a right-hand rule for \vec{v} \times \vec{B}:

  1. Point your fingers in the direction of \vec{v}.
  2. Curl them toward \vec{B} (the smaller angle between them).
  3. Your thumb points in the direction of \vec{F} for a positive charge.

If the charge is negative (like an electron), the force direction is opposite your thumb.

A very common exam error is to do the right-hand rule correctly but forget to flip the direction for negative charges.

Special cases you should recognize immediately

  • If \theta = 0 (particle moves parallel to field), then:

F = 0

  • If \theta = 90^\circ (perpendicular), then:

F = |q| v B

That perpendicular case is the one that gives pure circular motion.

Magnetic force does no work (in the ideal magnetic-only case)

Work depends on the component of force along the displacement. Since magnetic force is perpendicular to velocity, it does no work on the particle:

  • Speed stays constant (if only magnetic forces act).
  • Kinetic energy stays constant.

Students often expect the particle to “speed up toward the magnet,” but magnetism in this context mostly redirects motion rather than speeding it up.

Circular motion of a charged particle in a uniform magnetic field

If \vec{v} is perpendicular to a uniform \vec{B}, the magnetic force provides the centripetal force.

Set magnetic force equal to centripetal force:

|q|vB = \frac{mv^2}{r}

Solve for radius:

r = \frac{mv}{|q|B}

This relationship is extremely testable: bigger B means tighter circle (smaller r), larger mass means bigger circle.

The period (time for one full revolution) is:

T = \frac{2\pi r}{v}

Substitute the radius expression to get:

T = \frac{2\pi m}{|q|B}

Notice T does not depend on speed—this surprises many students. Faster particles make larger circles, but they travel that larger circle faster in exactly the right way to keep the period the same.

If the velocity has a component parallel to \vec{B}, the motion becomes a helix: circular motion from the perpendicular component and constant forward motion from the parallel component.

Crossed fields and velocity selection (common AP application)

If a charged particle moves through perpendicular electric and magnetic fields, it can travel straight if the forces balance.

Electric force magnitude:

F_E = |q|E

Magnetic force magnitude (assuming perpendicular):

F_B = |q|vB

Straight-line condition: F_E = F_B, so:

|q|E = |q|vB

Cancel |q|:

v = \frac{E}{B}

This is the principle of a velocity selector: only particles with speed v = E/B go straight; others are deflected.

Worked example: force direction and magnitude

A proton (charge q = +1.60\times 10^{-19}\ \text{C}) moves east at v = 3.0\times 10^6\ \text{m/s} through a uniform magnetic field of B = 0.20\ \text{T} directed north. Find the direction and magnitude of the magnetic force.

Step 1: Identify angle. East is perpendicular to north, so \theta = 90^\circ and \sin\theta = 1.

Step 2: Magnitude.

F = |q| v B

F = (1.60\times 10^{-19})(3.0\times 10^6)(0.20)

F = 9.6\times 10^{-14}\ \text{N}

Step 3: Direction. Use right-hand rule for \vec{v} \times \vec{B}.

  • \vec{v} east
  • \vec{B} north
  • east cross north points up (out of the horizontal plane)

Because the proton is positive, the force is upward.

Worked example: radius of curvature

An electron moves perpendicular to a uniform magnetic field with speed v = 2.0\times 10^7\ \text{m/s} in B = 0.050\ \text{T}. Find the radius of its path. Use m_e = 9.11\times 10^{-31}\ \text{kg} and |q| = 1.60\times 10^{-19}\ \text{C}.

Step 1: Use the circular motion radius formula.

r = \frac{mv}{|q|B}

Step 2: Substitute.

r = \frac{(9.11\times 10^{-31})(2.0\times 10^7)}{(1.60\times 10^{-19})(0.050)}

Compute numerator and denominator:

\text{numerator} = 1.822\times 10^{-23}

\text{denominator} = 8.0\times 10^{-21}

r = 2.28\times 10^{-3}\ \text{m}

So the electron’s path radius is about 2.3\ \text{mm}.

Exam Focus

  • Typical question patterns:
    • Use \vec{F} = q\vec{v} \times \vec{B} to find force direction (often with “into/out of page” diagrams).
    • Compute radius or period for circular motion in a uniform \vec{B}.
    • Apply crossed \vec{E} and \vec{B} fields to find the one speed that gives a straight path.
  • Common mistakes:
    • Forgetting \sin\theta and treating the force as nonzero when \vec{v} is parallel to \vec{B}.
    • Using the right-hand rule but not reversing direction for a negative charge.
    • Assuming magnetic force changes speed (it changes direction; speed stays constant if only magnetic forces act).

Magnetic Force on Current-Carrying Wires

Why a wire experiences a magnetic force

A current-carrying wire contains many moving charges. Each charge experiences a magnetic force, and the combined effect is a net force on the wire segment.

This idea is crucial because it connects microscopic physics (forces on charges) to macroscopic devices (motors, speakers, galvanometers). In many AP problems, you treat the wire as an object experiencing a force without tracking individual electrons.

The force on a straight wire segment in a uniform magnetic field

For a straight segment of wire of length L carrying current I in a uniform magnetic field \vec{B}:

\vec{F} = I\vec{L} \times \vec{B}

  • \vec{F} is the magnetic force on the wire segment
  • I is current (A)
  • \vec{L} is a vector pointing along the direction of conventional current, with magnitude equal to the length of the segment in the field (m)
  • \vec{B} is magnetic field (T)

Magnitude:

F = I L B \sin\theta

where \theta is the angle between the current direction (along \vec{L}) and \vec{B}.

Important conceptual parallels:

  • Compare F = I L B \sin\theta with F = |q| v B \sin\theta.
  • Both forces are maximal when motion/current is perpendicular to the field.
  • Both are zero when motion/current is parallel to the field.

Direction: right-hand rule for wires

Use the same cross product right-hand logic:

  1. Point fingers in direction of \vec{L} (current direction).
  2. Curl toward \vec{B}.
  3. Thumb gives \vec{F}.

A frequent confusion is mixing up “current direction” and “electron drift direction.” Again, use conventional current for right-hand rules.

Real-world meaning: how motors get a turning effect

A single straight wire in a magnetic field may translate, but a loop of wire can experience a torque (a tendency to rotate). This is the basic idea of an electric motor.

Magnetic torque on a current loop

For a loop with N turns and area A in a uniform field \vec{B}, the torque magnitude is:

\tau = N I A B \sin\theta

Here \theta is the angle between the loop’s area vector (perpendicular to the plane of the loop) and \vec{B}.

It’s common to define the magnetic dipole moment of the loop:

\mu = N I A

Then:

\tau = \mu B \sin\theta

This is not just a math convenience—\mu is a property of the loop that tells you how strongly it “wants” to align with a magnetic field (similar in spirit to how an electric dipole aligns with an electric field).

Force between parallel current-carrying wires

Two parallel wires carrying currents exert forces on each other because each wire produces a magnetic field that acts on the other wire.

For two long parallel wires separated by distance r, the force per unit length is:

\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi r}

  • If currents are in the same direction, the wires attract.
  • If currents are in opposite directions, the wires repel.

This result is a classic test of your ability to combine ideas:

  1. Find B from a long straight wire.
  2. Use F = I L B on the second wire.

Worked example: force on a wire segment

A straight wire segment of length L = 0.30\ \text{m} carries I = 4.0\ \text{A} through a uniform magnetic field of magnitude B = 0.50\ \text{T}. The current is perpendicular to the field. Find the force magnitude.

Step 1: Choose the formula.

F = I L B \sin\theta

Step 2: Use \theta = 90^\circ so \sin\theta = 1.

F = (4.0)(0.30)(0.50)

F = 0.60\ \text{N}

So the wire experiences a force of 0.60\ \text{N}.

Worked example: parallel wires (direction + magnitude)

Two long parallel wires are r = 0.040\ \text{m} apart. They carry currents I_1 = 6.0\ \text{A} and I_2 = 2.0\ \text{A} in the same direction. Find the force per meter between them.

Step 1: Use the force-per-length relation.

\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi r}

Step 2: Substitute.

\frac{F}{L} = \frac{(4\pi\times 10^{-7})(6.0)(2.0)}{2\pi(0.040)}

Cancel \pi and simplify:

\frac{F}{L} = \frac{(48\times 10^{-7})}{(0.080)}

\frac{F}{L} = 6.0\times 10^{-5}\ \text{N/m}

Step 3: Direction (attract or repel). Same current direction means attraction.

Common conceptual traps (woven into many problems)

  • “A wire is pushed toward the magnet.” Not generally. The force direction depends on current direction relative to \vec{B}; it can be up, down, into the page, etc.
  • Forgetting what L means in F = I L B \sin\theta. It’s the length of wire actually in the magnetic field region and oriented with a clear angle to \vec{B}.
  • Thinking a current loop always feels a net force. In a uniform \vec{B}, a closed loop often has forces that cancel for net force but produce a net torque.

Exam Focus

  • Typical question patterns:
    • Compute force magnitude/direction on a wire segment using \vec{F} = I\vec{L} \times \vec{B}.
    • Analyze a current loop in a uniform field to find torque and the orientation of stable equilibrium.
    • Combine B = \mu_0 I/(2\pi r) with F = I L B to get force between wires.
  • Common mistakes:
    • Using the wrong angle \theta (it’s between current direction and \vec{B}).
    • Confusing torque with force (a loop can have zero net force but nonzero torque).
    • Dropping vector direction reasoning and guessing “toward/away” instead of using a right-hand rule.