Magnetic Fields

Magnetic Fields and Forces

A magnetic field is a region in which a moving charge or current-carrying conductor experiences a force. It is generated either by permanent magnets or by moving electric charges.

Direction of Magnetic Force

For a moving charge, the direction of force is given by Fleming's Left Hand Rule:

• First finger → Magnetic field ($\vec{B}$)
• Second finger → Current ($\vec{I}$) or velocity of a positive charge ($\vec{v}$)
• Thumb → Force ($\vec{F}$)

For a negatively charged particle, reverse the direction of velocity.

Magnetic Force Equation (Lorentz Force)

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

• $\vec{F}$: Magnetic force (vector)
• $q$: Electric charge
• $\vec{v}$: Velocity of the particle
• $\vec{B}$: Magnetic field (magnetic flux density)

Magnitude of Magnetic Force

$$F = qvB \sin \theta$$

• $\theta$: Angle between $\vec{v}$ and $\vec{B}$
• Maximum force: $\theta = 90^\circ$
• No force: $\theta = 0^\circ$ or $180^\circ$

Unit of Magnetic Field (Tesla, T)

$$1 \, \text{T} = \frac{1 \, \text{N}}{1 \, \text{C} \cdot 1 \, \text{m/s}}$$

Magnetic Force on Moving Charges

Important Notes

1. The magnetic force is always perpendicular to both the velocity and magnetic field.
2. It does no work on the particle — it only changes its direction.
3. If $\vec{v} \perp \vec{B}$, the motion is circular.

Motion in Uniform Magnetic Field

The magnetic force acts as the centripetal force:

$$qvB = \frac{mv^2}{r} \Rightarrow r = \frac{mv}{qB}$$

Angular speed:

$$\omega = \frac{v}{r} = \frac{qB}{m}$$

Time period:

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

Note: Both $\omega$ and $T$ are independent of the radius and speed of the particle.

Motion at an Angle (Helical Path)

When a charged particle enters a magnetic field at an angle $\theta$, its velocity can be resolved into:

1. $v_x = v \cos \theta$: Parallel to the magnetic field → Uniform motion
2. $v_y = v \sin \theta$: Perpendicular to the field → Circular motion

Radius of the Helix

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

Pitch of the Helix (distance moved along field per cycle)

$$p = v \cos \theta \cdot T = v \cos \theta \cdot \frac{2\pi m}{qB}$$

Examples

Example 1

Given: $q = 3 \mu C$, $v = 5 \times 10^6 \, \text{m/s}$, $B = 0.08 \, \text{T}$

Solution:

$$F = qvB = 3 \times 10^{-6} \cdot 5 \times 10^6 \cdot 0.08 = 1.2 \, \text{N}$$

Example 2

Proton: $v = 8.5 \times 10^7 \, \text{m/s}$, $r = 0.68 \, \text{m}$

Constants: $q = 1.6 \times 10^{-19} \, \text{C}$, $m = 1.67 \times 10^{-27} \, \text{kg}$

Solution:

$$B = \frac{mv}{qr} = \frac{1.67 \times 10^{-27} \cdot 8.5 \times 10^7}{1.6 \times 10^{-19} \cdot 0.68} = 1.3 \, \text{T}$$

Example 3

Given: $\theta = 72^\circ$, $B = 0.32 \, \text{T}$, $v = 5.5 \times 10^6 \, \text{m/s}$

Use:

$$r = \frac{mv \sin \theta}{qB}, \quad p = v \cos \theta \cdot \frac{2\pi m}{qB}$$