Force on a Moving Charge in a Uniform Magnetic Field

A charged particle moving through a region of magnetic field experiences a deflecting force called the magnetic force. Unlike electric force, this force acts only when the charge is in motion and always acts perpendicular to the velocity, causing the particle to change direction without changing its speed.

Think of the magnetic force like a string pulling a ball in a circle — it constantly changes the ball's direction (centripetal effect) but never speeds it up or slows it down, since the pull is always sideways to the motion.

Scalar Form: F = qvB sin ⁡θ

Vector Form: \[\vec F\] = q(\[\vec v\] × \[\vec B\])

Derivation logic: The magnetic force is defined empirically such that it is proportional to the charge, the speed of the charge, the magnetic field strength, and the sine of the angle between v and B. Since the cross product \[\vec v\] × \[\vec B\] naturally produces a vector of magnitude vB sin ⁡θ directed perpendicular to both v and B, this matches experimental observation exactly, giving \[\vec F\] = q(\[\vec v\] × \[\vec B\]).

Direction Rule: Point fingers along v, curl them toward B; the thumb points along F (for positive charge). For a negative charge, reverse the direction.

When both electric and magnetic fields act simultaneously on a moving charge:

This combined expression is called the Lorentz force and forms the basis of devices like the cyclotron and velocity selector.

Problem: A proton moves with velocity \[\vec v\] = 2 × 105\[\hat i\] m/s in a magnetic field \[\vec B\] = (2\[\hat i\] + 3\[\hat j\]) T. Find the magnetic force on the proton. (q = 1.6 × 10^-19 C)

Solution:

Since \[\hat i\] × \[\hat i\] = 0 and \[\hat i\] × \[\hat j\] = \[\hat k\]:

Answer: \[\vec F\] = 9.6 × 10⁻¹⁴\[\hat k\] N