Magnetic Fields and Forces

Lorentz force, motion of charges in magnetic fields

Magnetic Fields and Forces

Magnetic Force on Moving Charge

Lorentz force: $\vec{F} = q\vec{v} \times \vec{B}$

Magnitude: $F = qvB\sin\theta$

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

Direction: Right-hand rule

Properties:

Force perpendicular to both $\vec{v}$ and $\vec{B}$
Magnetic force does no work ( $\vec{F} \perp \vec{v}$ )
Changes direction, not speed

Circular Motion in Magnetic Field

For $\vec{v} \perp \vec{B}$ :

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

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

Period: $T = \frac{2\pi r}{v} = \frac{2\pi m}{qB}$

Cyclotron frequency: $f = \frac{qB}{2\pi m}$

(Independent of $v$ and $r$ !)

Helical Motion

If $\vec{v}$ has components both parallel and perpendicular to $\vec{B}$ :

Perpendicular component causes circular motion
Parallel component causes uniform motion
Result: helix

Pitch of helix: $p = v_{\parallel}T = \frac{2\pi m v_{\parallel}}{qB}$

Combined Electric and Magnetic Fields

Total force: $\vec{F} = q(\vec{E} + \vec{v} \times \vec{B})$

Velocity selector:

Crossed $\vec{E}$ and $\vec{B}$ perpendicular to $\vec{v}$ :

Particles with $v = E/B$ pass straight through ( $\vec{F} = 0$ ).

Magnetic Force on Current

Current $I$ in wire of length $\vec{L}$ in field $\vec{B}$ :

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

For straight wire: $F = ILB\sin\theta$

For curved wire: $\vec{F} = \int I \, d\vec{l} \times \vec{B}$

Torque on Current Loop

Rectangular loop with area $A$ , current $I$ , in uniform field $\vec{B}$ :

Magnetic dipole moment: $\vec{\mu} = IA\hat{n}$

(where $\hat{n}$ is normal to loop, by right-hand rule)

Torque: $\vec{\tau} = \vec{\mu} \times \vec{B}$

$\tau = \mu B\sin\theta = IAB\sin\theta$

Potential energy: $U = -\vec{\mu} \cdot \vec{B} = -\mu B\cos\theta$

(Minimum when $\vec{\mu}$ parallel to $\vec{B}$ )

Applications

Mass spectrometer: Separates ions by mass using circular motion in $\vec{B}$

Cyclotron: Accelerates particles using constant frequency RF field

Hall effect: Voltage across conductor perpendicular to current and $\vec{B}$

$V_H = \frac{IB}{nqt}$

where $n$ is charge carrier density, $t$ is thickness.

📚 Practice Problems

1Problem 1easy

❓ Question:

A proton (q = 1.6 × 10⁻¹⁹ C, m = 1.67 × 10⁻²⁷ kg) enters a uniform magnetic field B = 0.5 T with velocity v = 2.0 × 10⁶ m/s perpendicular to the field. Find: (a) the magnitude of the magnetic force, (b) the radius of the circular path, and (c) the period of the motion.

💡 Show Solution

Given:

q = 1.6 × 10⁻¹⁹ C
m = 1.67 × 10⁻²⁷ kg
B = 0.5 T
v = 2.0 × 10⁶ m/s
θ = 90° (perpendicular)

(a) Magnetic force:

$F = qvB\sin\theta = (1.6 \times 10^{-19})(2.0 \times 10^6)(0.5)(1)$

$F = \boxed{1.6 \times 10^{-13} \text{ N}}$

(b) Radius of path:

Magnetic force provides centripetal force: $qvB = \frac{mv^2}{r}$

$r = \frac{mv}{qB} = \frac{(1.67 \times 10^{-27})(2.0 \times 10^6)}{(1.6 \times 10^{-19})(0.5)}$

$r = \boxed{4.18 \times 10^{-2} \text{ m} = 4.18 \text{ cm}}$

(c) Period:

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

$T = \frac{2\pi (1.67 \times 10^{-27})}{(1.6 \times 10^{-19})(0.5)}$

$T = \boxed{1.31 \times 10^{-7} \text{ s} = 131 \text{ ns}}$

Note: Period is independent of velocity!

2Problem 2medium

❓ Question:

A rectangular current loop (dimensions: 0.2 m × 0.3 m) carries current I = 5.0 A. The loop is placed in a uniform magnetic field B = 0.8 T with the field parallel to the plane of the loop, making angle θ = 30° with the normal to the loop. Find: (a) the magnetic dipole moment, (b) the torque on the loop, and (c) the orientation angle for maximum torque.

💡 Show Solution

Given:

Dimensions: 0.2 m × 0.3 m
I = 5.0 A
B = 0.8 T
θ = 30° (angle between normal and field)

(a) Magnetic dipole moment:

$\mu = IA = I(\text{length} \times \text{width})$

$\mu = (5.0)(0.2)(0.3) = \boxed{0.3 \text{ A·m}^2}$

Direction: perpendicular to loop (right-hand rule)

(b) Torque:

$\tau = \mu B \sin\theta = (0.3)(0.8)\sin(30°)$

$\tau = (0.24)(0.5) = \boxed{0.12 \text{ N·m}}$

Direction: tends to align μ with B

(c) Maximum torque:

Torque is maximum when $\sin\theta = 1$ , i.e., when:

$\theta = \boxed{90°}$

This occurs when the normal to the loop is perpendicular to B, meaning the magnetic field is in the plane of the loop.

Maximum torque: $\tau_{max} = \mu B = (0.3)(0.8) = 0.24 \text{ N·m}$

3Problem 3hard

❓ Question:

A particle with charge q = 2.0 μC and mass m = 1.0 × 10⁻⁸ kg enters a region with perpendicular electric field E = 500 V/m and magnetic field B = 0.1 T. The particle moves in a straight line. Find: (a) the velocity of the particle (velocity selector), (b) if E is then turned off, find the radius of the resulting circular path, and (c) explain how this setup could be used as a mass spectrometer.

💡 Show Solution

Given:

q = 2.0 μC = 2.0 × 10⁻⁶ C
m = 1.0 × 10⁻⁸ kg
E = 500 V/m
B = 0.1 T
Straight-line motion initially

(a) Velocity (velocity selector):

For straight-line motion, electric and magnetic forces balance: $F_E = F_B$ $qE = qvB$

$v = \frac{E}{B} = \frac{500}{0.1}$

$v = \boxed{5000 \text{ m/s}}$

This velocity is independent of q and m!

(b) Radius with E off:

With only magnetic force: $r = \frac{mv}{qB} = \frac{(1.0 \times 10^{-8})(5000)}{(2.0 \times 10^{-6})(0.1)}$

$r = \frac{5.0 \times 10^{-5}}{2.0 \times 10^{-7}}$

$r = \boxed{0.25 \text{ m} = 25 \text{ cm}}$

(c) Mass spectrometer principle:

Since $r = \frac{mv}{qB}$ and $v = E/B$ :

$r = \frac{m(E/B)}{qB} = \frac{mE}{qB^2}$

For particles with same q but different m: $\frac{r_1}{r_2} = \frac{m_1}{m_2}$

How it works:

Velocity selector ensures all particles have same v (regardless of m)
Magnetic field separates particles by mass
Heavier particles follow larger radius paths
Detector measures radius → determines mass
Can separate isotopes (same q, different m)

Example: ²³⁵U and ²³⁸U ions would separate into different paths.

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