Magnetic Force

Overview

Magnetic fields exert forces on moving charges and current-carrying wires. These forces are fundamental to the operation of motors, generators, and many other devices.

Force on a Moving Charge

Lorentz Force (Magnetic Part)

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

Magnitude:

$$F = qvB \sin(\theta)$$

Where $\theta$ is the angle between $\vec{v}$ and $\vec{B}$

Properties

• Force is perpendicular to both $\vec{v}$ and $\vec{B}$
• No work done (speed unchanged)
• Direction given by right-hand rule
• Maximum when $\vec{v} \perp \vec{B}$
• Zero when $\vec{v} \parallel \vec{B}$

Right-Hand Rule for Forces

1. Point fingers in direction of $\vec{v}$
2. Curl toward $\vec{B}$
3. Thumb points in direction of $\vec{F}$ (for positive charge)
4. For negative charge, force is opposite

Motion of Charged Particles

In Uniform Magnetic Field

$\vec{v} \perp \vec{B}$: Circular Motion

$$r = \frac{mv}{qB}$$

Period (independent of speed):

$$T = \frac{2\pi m}{qB}$$

Cyclotron frequency:

$$f = \frac{qB}{2\pi m}$$

$\vec{v}$ at angle to $\vec{B}$: Helical Motion

• Circular motion perpendicular to $\vec{B}$
• Constant velocity parallel to $\vec{B}$
• Results in spiral path

Radius and Momentum

$$r = \frac{mv}{qB} = \frac{p}{qB}$$

Force on Current-Carrying Wire

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

Magnitude:

$$F = BIL \sin(\theta)$$

Where:

• $I$ = current
• $L$ = length of wire in field
• $\theta$ = angle between wire and $\vec{B}$

Direction: right-hand rule (fingers = $I$, curl toward $\vec{B}$, thumb = $\vec{F}$)

Force Between Parallel Wires

Two parallel wires separated by distance $d$:

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

• Currents in same direction: attractive
• Currents in opposite direction: repulsive

This defines the ampere: 1 A produces $2 \times 10^{-7}$ N/m between wires 1 m apart.

Torque on Current Loop

A rectangular loop (area $A$) in a uniform field:

$$\tau = NIAB \sin(\theta)$$ $$\vec{\tau} = \vec{\mu} \times \vec{B}$$

Where:

• $N$ = number of turns
• $\mu = NIA$ = magnetic dipole moment
• $\theta$ = angle between $\vec{\mu}$ and $\vec{B}$

Maximum Torque

$$\tau_{\max} = NIAB \quad \text{(when } \theta = 90°\text{)}$$

Hall Effect

When current flows through a conductor in a magnetic field:

• Charges are deflected to one side
• Creates a transverse voltage (Hall voltage)

$$V_H = Bvd = \frac{BId}{nqA}$$

Applications: measure $B$, determine sign of charge carriers

Examples

Example 1: Force on Electron

An electron ($v = 3 \times 10^6$ m/s) moves perpendicular to $B = 0.5$ T.

$$F = qvB = 1.6 \times 10^{-19} \times 3 \times 10^6 \times 0.5 = 2.4 \times 10^{-13} \text{ N}$$

Example 2: Circular Motion

A proton ($m = 1.67 \times 10^{-27}$ kg) moves at $2 \times 10^6$ m/s perpendicular to $B = 0.1$ T.

$$r = \frac{mv}{qB} = \frac{1.67 \times 10^{-27} \times 2 \times 10^6}{1.6 \times 10^{-19} \times 0.1}$$ $$r = 0.209 \text{ m} = 20.9 \text{ cm}$$

Period:

$$T = \frac{2\pi m}{qB} = \frac{2\pi \times 1.67 \times 10^{-27}}{1.6 \times 10^{-19} \times 0.1}$$ $$T = 6.56 \times 10^{-7} \text{ s} = 656 \text{ ns}$$

Example 3: Force on Wire

A 50 cm wire carrying 8 A is perpendicular to $B = 0.3$ T.

$$F = BIL = 0.3 \times 8 \times 0.5 = 1.2 \text{ N}$$

Example 4: Force Between Wires

Two parallel wires 5 cm apart carry 10 A each (same direction). Find force per unit length.

$$\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d} = \frac{4\pi \times 10^{-7} \times 10 \times 10}{2\pi \times 0.05}$$ $$\frac{F}{L} = 4 \times 10^{-4} \text{ N/m (attractive)}$$

Example 5: Motor Torque

A rectangular coil (20 turns, 10 cm × 5 cm) carries 2 A in $B = 0.8$ T. Find maximum torque.

$$A = 0.1 \times 0.05 = 0.005 \text{ m}^2$$ $$\tau_{\max} = NIAB = 20 \times 2 \times 0.005 \times 0.8 = 0.16 \text{ N·m}$$

Example 6: Mass Spectrometer

An ion ($q = e$, $m$ = unknown) is accelerated through 1000 V and enters $B = 0.5$ T, making a semicircle of radius 10 cm. Find $m$.

Velocity from energy:

$$\frac{1}{2}mv^2 = qV \to v = \sqrt{\frac{2qV}{m}}$$

From radius:

$$r = \frac{mv}{qB}$$ $$m = \frac{qBr}{v} = \frac{qBr}{\sqrt{2qV/m}}$$ $$m^2 = \frac{q^2 B^2 r^2 m}{2qV} = \frac{qB^2 r^2}{2V}$$ $$m = \frac{qB^2 r^2}{2V} = \frac{1.6 \times 10^{-19} \times 0.25 \times 0.01}{2 \times 1000}$$ $$m = 2 \times 10^{-25} \text{ kg} \approx 120 \text{ amu (like Sn ion)}$$

Applications

• Electric motors: Torque on current loops
• Mass spectrometers: Separating ions by mass
• Cyclotrons: Accelerating particles
• Hall sensors: Measuring magnetic fields
• Loudspeakers: Force on coil in magnetic field