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Magnetic force on charges and currents explains how moving charges, wires, and current loops interact with magnetic fields. Students need this cheat sheet to connect vector directions, formulas, and real physical motion in one place. It is especially useful for problems involving particles in fields, mass spectrometers, motors, and current-carrying wires. The most important idea is that magnetic force acts only on moving charges or currents and depends on the component of motion perpendicular to the magnetic field. For a single charge, the force magnitude is FB=qvBsinθF_B = |q|vB\sin\theta, and its direction comes from the right-hand rule with charge sign included. In a uniform magnetic field, perpendicular motion can produce circular motion with r=mvqBr = \frac{mv}{|q|B}. For currents, the main formulas are FB=ILBsinθF_B = ILB\sin\theta for a straight wire and τ=NIABsinθ\tau = NIAB\sin\theta for a current loop.

Key Facts

  • The magnetic force on a moving charge has magnitude FB=qvBsinθF_B = |q|vB\sin\theta, where θ\theta is the angle between v\vec{v} and B\vec{B}.
  • The vector form of magnetic force on a charge is FB=qv×B\vec{F}_B = q\vec{v} \times \vec{B}, so a negative charge feels force opposite the right-hand-rule direction.
  • A charge moving parallel or antiparallel to a magnetic field feels no magnetic force because sin0=0\sin 0^{\circ} = 0 and sin180=0\sin 180^{\circ} = 0.
  • A charge moving perpendicular to a uniform magnetic field follows circular motion when magnetic force supplies centripetal force, so qvB=mv2r|q|vB = \frac{mv^2}{r}.
  • The radius of circular motion in a uniform magnetic field is r=mvqBr = \frac{mv}{|q|B}.
  • The angular speed and period for perpendicular circular motion are ω=qBm\omega = \frac{|q|B}{m} and T=2πmqBT = \frac{2\pi m}{|q|B}.
  • The magnetic force on a straight current-carrying wire is FB=ILBsinθF_B = ILB\sin\theta, where LL is the wire length inside the magnetic field.
  • The torque on a current loop is τ=NIABsinθ\tau = NIAB\sin\theta, where NN is the number of turns and AA is the loop area.

Vocabulary

Magnetic field
A region described by B\vec{B} where moving charges or currents can experience magnetic force.
Lorentz force
The total electromagnetic force on a charge, often written F=qE+qv×B\vec{F} = q\vec{E} + q\vec{v} \times \vec{B}.
Right-hand rule
A direction rule where fingers follow v\vec{v} or current, curl toward B\vec{B}, and the thumb gives the force direction for positive charge or conventional current.
Centripetal force
The inward net force needed for circular motion, with magnitude Fc=mv2rF_c = \frac{mv^2}{r}.
Conventional current
The direction of current defined as the motion of positive charge, even though electrons in metals move the opposite way.
Magnetic torque
The turning effect on a current loop in a magnetic field, with magnitude τ=NIABsinθ\tau = NIAB\sin\theta.

Common Mistakes to Avoid

  • Using FB=qvBF_B = qvB for every angle is wrong because the correct magnitude is FB=qvBsinθF_B = |q|vB\sin\theta and only the perpendicular component matters.
  • Forgetting the sign of the charge gives the wrong force direction because FB=qv×B\vec{F}_B = q\vec{v} \times \vec{B} reverses direction for negative charges.
  • Thinking magnetic force changes speed is wrong for a charge in a magnetic field alone because FB\vec{F}_B is perpendicular to v\vec{v} and changes direction, not kinetic energy.
  • Using the full wire length when only part is in the field is wrong because FB=ILBsinθF_B = ILB\sin\theta uses the length LL inside the magnetic field.
  • Confusing the angle in torque problems is wrong because τ=NIABsinθ\tau = NIAB\sin\theta uses the angle between the loop’s area vector and B\vec{B}, not necessarily the plane of the loop.

Practice Questions

  1. 1 A proton with charge q=1.60×1019Cq = 1.60 \times 10^{-19}\,\text{C} moves at v=2.0×106m/sv = 2.0 \times 10^6\,\text{m/s} perpendicular to a magnetic field of B=0.30TB = 0.30\,\text{T}. What is the magnetic force magnitude?
  2. 2 An electron moves perpendicular to a uniform magnetic field of B=0.050TB = 0.050\,\text{T} with speed v=3.0×106m/sv = 3.0 \times 10^6\,\text{m/s}. Using me=9.11×1031kgm_e = 9.11 \times 10^{-31}\,\text{kg} and qe=1.60×1019C|q_e| = 1.60 \times 10^{-19}\,\text{C}, find the radius of its circular path.
  3. 3 A straight wire of length L=0.75mL = 0.75\,\text{m} carries current I=4.0AI = 4.0\,\text{A} at 9090^{\circ} to a magnetic field of B=0.20TB = 0.20\,\text{T}. Find the force on the wire.
  4. 4 A charged particle enters a uniform magnetic field with velocity exactly parallel to B\vec{B}. Explain why its path does not curve due to the magnetic field.