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.

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

(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!