← General Physics II
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Magnetic Fields & Forces

Magnetism and electricity are two faces of the same force. A magnetic field exerts a force on any moving charge — perpendicular to both the velocity and the field. This seemingly simple rule explains electric motors, mass spectrometers, the aurora borealis, and the confinement of plasma in fusion reactors.

Key Concepts

Magnetic Field and Force on a Moving Charge
A charge qq moving with velocity v\vec{v} in magnetic field B\vec{B} experiences force F=qv×B\vec{F} = q\vec{v}\times\vec{B}. Magnitude: F=qvBsinθF = |q|vB\sin\theta, where θ\theta is the angle between v\vec{v} and B\vec{B}. The force is always perpendicular to v\vec{v}, so it changes direction but never speed or kinetic energy.
Circular Motion in a Magnetic Field
When a charged particle moves perpendicular to a uniform B\vec{B}, the magnetic force provides centripetal acceleration. The orbital radius is r=mv/(qB)r = mv/(|q|B) and the period is T=2πm/(qB)T = 2\pi m/(|q|B) — independent of speed (cyclotron principle).
Magnetic Force on a Current-Carrying Wire
A straight wire of length LL carrying current II in field B\vec{B} experiences force F=BILsinθF = BIL\sin\theta. Direction from the right-hand rule (or F=IL×B\vec{F} = I\vec{L}\times\vec{B}).
Torque on a Current Loop
A flat current loop of NN turns, area AA, carrying current II in field BB experiences torque τ=NIABsinθ\tau = NIAB\sin\theta, where θ\theta is the angle between the loop's magnetic dipole moment μ=NIAn^\vec{\mu} = NIA\hat{n} and B\vec{B}. This is the operating principle of galvanometers and electric motors.
Hall Effect
When a current-carrying conductor is placed in a perpendicular magnetic field, charge carriers are deflected sideways, building up a transverse voltage (the Hall voltage). The sign of the Hall voltage reveals whether carriers are positive or negative.

Key Equations

Magnetic force on a charge
F=qvBsinθF = |q|vB\sin\theta
Magnitude of force on charge q moving at speed v in field B; θ is the angle between v and B. Zero if parallel.
Radius of circular orbit
r=mvqBr = \frac{mv}{|q|B}
Radius of circular path for a particle of mass m and charge q moving perpendicular to field B.
Force on a current-carrying wire
F=BILsinθF = BIL\sin\theta
Force on a straight wire of length L carrying current I in field B; θ between wire and B.
Torque on a current loop
τ=NIABsinθ\tau = NIAB\sin\theta
Torque on a loop of N turns, area A, current I in field B; θ between magnetic moment and B.
Worked Example

Proton in a Uniform Magnetic Field

Problem

A proton (m=1.673×1027m = 1.673\times10^{-27} kg, q=1.602×1019q = 1.602\times10^{-19} C) enters a region of uniform B=0.50B = 0.50 T with speed v=3.0×106v = 3.0\times10^6 m/s perpendicular to the field. Find (a) the radius of its circular orbit and (b) its orbital period.

Solution

(a) The magnetic force provides centripetal force: qvB=mv2/r|q|vB = mv^2/r, so:

r=mvqB=(1.673×1027)(3.0×106)(1.602×1019)(0.50)=5.02×10218.01×1020=0.0626 m6.26 cmr = \frac{mv}{|q|B} = \frac{(1.673\times10^{-27})(3.0\times10^6)}{(1.602\times10^{-19})(0.50)} = \frac{5.02\times10^{-21}}{8.01\times10^{-20}} = 0.0626\text{ m} \approx 6.26\text{ cm}

(b) Period (circumference divided by speed):

T=2πrv=2π(0.0626)3.0×106=1.31×107 s131 nsT = \frac{2\pi r}{v} = \frac{2\pi(0.0626)}{3.0\times10^6} = 1.31\times10^{-7}\text{ s} \approx 131\text{ ns}

Note: T=2πm/(qB)T = 2\pi m/(|q|B) is independent of speed — this is the cyclotron principle.

Answer r ≈ 6.26 cm; T ≈ 131 ns.
Practice

Exercises

7 problems
1 of 7

A charge q=+2μq = +2\,\muC moves at v=3.0×105v = 3.0\times10^5 m/s perpendicular to a field B=0.40B = 0.40 T. What is the magnitude of the magnetic force (in N) on it?

N
2 of 7

A straight wire carries I=5.0I = 5.0 A and has length L=0.20L = 0.20 m in a uniform field B=0.80B = 0.80 T perpendicular to the wire. What is the force (in N) on the wire?

N
3 of 7

A proton (m=1.673×1027m = 1.673\times10^{-27} kg, q=1.602×1019q = 1.602\times10^{-19} C) moves at v=2.0×106v = 2.0\times10^6 m/s perpendicular to B=0.30B = 0.30 T. What is the radius (in cm) of its circular orbit?

cm
4 of 7

A charge q=+3μq = +3\,\muC moves at v=4.0×104v = 4.0\times10^4 m/s at θ=30°\theta = 30° to a field B=2.0B = 2.0 T. What is the magnitude of the magnetic force (in N)?

N
5 of 7

A rectangular coil has N=100N = 100 turns, area A=0.020A = 0.020 m², carries I=0.50I = 0.50 A, and sits in a field B=0.80B = 0.80 T. What is the maximum torque (in N·m) on the coil?

N·m
6 of 7

An electron (m=9.11×1031m = 9.11\times10^{-31} kg, q=1.602×1019q = 1.602\times10^{-19} C) moves at v=5.0×106v = 5.0\times10^6 m/s perpendicular to B=0.10B = 0.10 T. What is the radius (in mm) of its circular orbit?

mm
7 of 7

A proton (m=1.673×1027m = 1.673\times10^{-27} kg, q=1.602×1019q = 1.602\times10^{-19} C) travels in a circular orbit of radius r=5.0r = 5.0 cm in a field B=0.50B = 0.50 T. What is its speed (in Mm/s, i.e. 10610^6 m/s)?

Mm/s

Key Takeaways

  • The magnetic force F=qv×B\vec{F} = q\vec{v}\times\vec{B} is always perpendicular to velocity — it curves paths but never does work.
  • Use the right-hand rule: fingers point along v\vec{v}, curl toward B\vec{B}, thumb points in the direction of F\vec{F} for positive charge.
  • Circular orbit radius r=mv/(qB)r = mv/(|q|B): heavier or faster particles curve less; stronger field curves them more.
  • The cyclotron period T=2πm/(qB)T = 2\pi m/(|q|B) is speed-independent — the basis of particle accelerators.
  • A current loop in a field experiences torque that aligns its magnetic moment with B\vec{B} — the same principle as a compass needle.