Chapter 21: Magnetic Forces and Magnetic Fields

21.1 Magnetic Fields

A compass needle is a permanent magnet with a north (N) and south (S) magnetic pole.
Magnetic poles behave similarly to electric charges: like poles repel, and unlike poles attract.
Magnetic poles cannot be isolated; they always exist in pairs.
A magnetic field surrounds a magnet. The direction of the field at a point is the direction indicated by the north pole of a compass needle at that point.
The Earth has a magnetic field with a north magnetic pole and a south magnetic pole. The north magnetic pole is near the Earth's geographic north pole.

21.2 The Force That a Magnetic Field Exerts on a Charge

A charge placed in an electric field experiences a force: $F_E = qE$
For a charge to experience a magnetic force, two conditions must be met:
- The charge must be moving.
- The velocity of the charge must have a component perpendicular to the magnetic field.
The magnetic force $F$ on a moving charge $q$ in a magnetic field $B$ is given by: $F = qvB\sin\theta$ , where $\theta$ is the angle between the velocity and the magnetic field.
The units of magnetic field $B$ are Tesla (T), where 1 T = 1 Wb/m² = 10⁴ Gauss (G).
Right Hand Rule No. 1 (RHR-1):
- Point fingers of the right hand along the magnetic field direction.
- Point the thumb along the velocity of the charge.
- The palm faces the direction of the magnetic force on a positive charge. The force on a negative charge is opposite to this direction.
The magnitude of the magnetic field is defined as: $B = \frac{F}{qv \sin \theta}$ where $\theta$ is between 0 and 180 degrees.
Units of magnetic field: $1 \text{ tesla (T)} = \frac{\text{newton} \cdot \text{second}}{\text{coulomb} \cdot \text{meter}} = \frac{\text{N} \cdot \text{s}}{\text{C} \cdot \text{m}}$
$1 \text{ gauss} = 10^{-4} \text{ tesla}$

Example: Magnetic Forces on Charged Particles

A proton with a speed of $5.0 \times 10^6 \text{ m/s}$ enters a 0.40 T magnetic field at an angle of 30.0 degrees.
- The magnitude of the force on the proton is: $F = qvB \sin \theta = (1.60 \times 10^{-19} \text{ C})(5.0 \times 10^6 \text{ m/s})(0.40 \text{ T}) \sin 30.0^{\circ} = 1.6 \times 10^{-13} \text{ N}$
- The acceleration of the proton is: $a = \frac{F}{m} = \frac{1.6 \times 10^{-13} \text{ N}}{1.67 \times 10^{-27} \text{ kg}} = 9.6 \times 10^{13} \text{ m/s}^2$
- If the particle were an electron, the magnitude of the force would be the same, but the direction would be opposite.
- The acceleration of the electron would be: $a = \frac{F}{m} = \frac{1.6 \times 10^{-13} \text{ N}}{9.11 \times 10^{-31} \text{ kg}} = 1.8 \times 10^{17} \text{ m/s}^2$

21.3 The Motion of a Charged Particle in a Magnetic Field

Charged Particle in a magnetic field vs electric field
A velocity selector uses electric and magnetic fields to measure the velocity of a charged particle by balancing the forces.
Electric force can do work on a charged particle, but magnetic force cannot because it's always perpendicular to the velocity.
The magnetic force is always perpendicular to the velocity, causing the particle to move in a circle.
The centripetal force is given by $F_c = m \frac{v^2}{r}$ . The magnetic force provides this centripetal force: $qvB = m \frac{v^2}{r}$ , so the radius of the circular path is $r = \frac{mv}{qB}$ .

Conceptual Example: Particle Tracks in a Bubble Chamber

Gamma rays transform into charged particles, producing spiral tracks in a bubble chamber due to a magnetic field. Determine the sign and speed of each particle based on track curvature.

21.4 The Mass Spectrometer

For a singly ionized particle starting from rest, the mass is given by: $m = \frac{q B^2 r^2}{2V}$
A mass spectrometer separates ions based on their mass-to-charge ratio.
Isotopes of an element can be identified using a mass spectrometer.

21.5 The Force on a Current in a Magnetic Field

The magnetic force on moving charges in a wire pushes the wire.
The force on a single charge is $F = qvB \sin \theta$ .
If a wire of length $L$ carries a current $I$ in a magnetic field $B$ , the magnetic force on the wire is $F = ILB \sin \theta$ .

Example: The Force and Acceleration in a Loudspeaker

A voice coil in a speaker with a diameter of 0.025 m, 55 turns, in a 0.10 T field carries a 2.0 A current.
- The magnetic force on the coil is: $F = N I L B \sin \theta = 55 (2.0 \text{ A}) (\pi \cdot 0.025 \text{ m}) (0.10 \text{ T}) \sin 90^{\circ} = 0.86 \text{ N}$
- If the mass of the coil and cone is 0.0200 kg, the acceleration is: $a = \frac{F}{m} = \frac{0.86 \text{ N}}{0.020 \text{ kg}} = 43 \text{ m/s}^2$

21.6 The Torque on a Current-Carrying Coil

The forces on a current-carrying loop in a magnetic field have equal magnitude but opposite directions, creating a torque.
The loop tends to rotate so its normal aligns with the magnetic field.
The net torque on a coil is given by: $\tau = IAB \sin \phi$ , where $\phi$ is the angle between the normal to the coil and the magnetic field.

Example: The Torque Exerted on a Current-Carrying Coil

A coil with area $2.0 \times 10^{-4} \text{ m}^2$ , 100 turns, and a 0.045 A current is in a 0.15 T magnetic field.
- The magnetic moment of the coil is: $NIA = (100)(0.045 \text{ A})(2.0 \times 10^{-4} \text{ m}^2) = 9.0 \times 10^{-4} \text{ A} \cdot \text{m}^2$
- The maximum torque is: $\tau = NIAB \sin \phi = (100)(0.045 \text{ A})(2.0 \times 10^{-4} \text{ m}^2)(0.15 \text{ T}) \sin 90^{\circ} = 1.4 \times 10^{-5} \text{ N} \cdot \text{m}$
In a DC motor, a split-ring commutator ensures continuous rotation by reversing the current direction.

21.7 Magnetic Fields Produced by Currents

Right-Hand Rule No. 2: Curl fingers in a half-circle. Point the thumb in the direction of current, then the finger tips point in direction of magnetic field.
The magnetic field around a long, straight wire is: $B = \frac{\mu0 I}{2 \pi r}$ , where $\mu0$ is the permeability of free space ( $4 \pi \times 10^{-7} \text{ T} \cdot \text{m/A}$ ).

Example: A Current Exerts a Magnetic Force on a Moving Charge

A wire carries 3.0 A. A particle with charge $6.5 \times 10^{-6} \text{ C}$ moves parallel to the wire at 0.050 m with a speed of 280 m/s.
- $\mu_0$ is permeability of free space.
The magnetic force is $F = qvB\sin\theta=qv \frac{\mu_0 I}{2 \pi r} \sin \theta$ .
Current-carrying wires exert forces on each other.

Conceptual Example: The Net Force That a Current-Carrying Wire Exerts on a Current Carrying Coil

Determine whether a coil is attracted to or repelled by a wire based on current directions.
The magnetic field at the center of a circular loop is: $B = \frac{\mu_0 I}{2R}$ .

Example: Finding the Net Magnetic Field

A wire carries 8.0 A, and a loop of radius 0.030 m carries 2.0 A. Find the magnetic field at the loop's center.
$B = \frac{\mu0}{2 \pi} \frac{I}{r}$ . $B = \frac{\mu0 I}{2R}$
$B = \frac{\mu0}{2} ( \frac{I1}{\pi r} - \frac{I_2}{R} ) = \frac{4 \pi \times 10^{-7}}{2 \pi} ( \frac{8}{0.03} - \frac{2}{0.03} )=1.1 \times 10^{-5} T$
The field lines around a bar magnet resemble those around a loop.
A solenoid consists of many loops of wire.

21.8 Ampere’s Law

Ampere's Law relates the integral of the magnetic field around a closed loop to the current passing through the loop.
For static magnetic fields, Ampère’s law states: $\oint B \cdot ds = \mu_0 I$ , where $I$ is the total steady current through the surface bounded by the closed path.
$\mu_0 = 4 \pi \times 10^{-7} \text{ T} \cdot \text{m/A}$ (permeability of free space).

21.9 Magnetic Materials

Electron “spin” and orbital motion give rise to magnetic properties.
Ferromagnetic materials have magnetic domains where electron spins are aligned.
Soft magnetic materials (e.g., iron) are easily magnetized but lose magnetism easily; hard magnetic materials (e.g., cobalt, nickel) are hard to magnetize but retain magnetism.
MRI (magnetic resonance imaging) uses magnetic fields to create detailed body images without X-ray risks.
Hydrogen atoms in the body emit radio waves in a magnetic field, which are used to create a map of their distribution.
MRI is a powerful diagnostic tool used by surgeons for live imaging during operations.