describe the motion of a charged particle moving in a uniform magnetic field perpendicular to the direction of motion of the particle

Force on a Current‑Carrying Conductor

Learning Objective

Describe the motion of a charged particle moving in a uniform magnetic field perpendicular to its velocity, and relate this motion to the magnetic force on a straight current‑carrying conductor.

Key Concepts

• Magnetic force on a moving charge (Lorentz force)
• Uniform circular motion when the velocity is perpendicular to the magnetic field
• Relation between charge motion and electric current
• Magnetic force on a current‑carrying wire

Magnetic Force on a Single Charge

The Lorentz force on a charge \$q\$ moving with velocity \$\mathbf{v}\$ in a magnetic field \$\mathbf{B}\$ is

\$\$

\mathbf{F}=q\,\mathbf{v}\times\mathbf{B}

\$\$

If \$\mathbf{v}\perp\mathbf{B}\$, the magnitude simplifies to \$F = qvB\$, and the direction is given by the right‑hand rule.

Circular Motion of a Charged Particle

Because the magnetic force is always perpendicular to the velocity, it provides the centripetal force required for circular motion.

\$\$

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

\$\$

Hence the radius of the trajectory is

\$\$

r = \frac{mv}{qB}

\$\$

and the angular (cyclotron) frequency is

\$\$

\omega = \frac{v}{r}= \frac{qB}{m}

\$\$

From a Single Charge to a Current‑Carrying Conductor

Consider a straight wire of length \$L\$ carrying a current \$I\$ placed in a uniform magnetic field \$\mathbf{B}\$, with the field perpendicular to the wire.

The current is the net charge flow per unit time:

\$\$

I = n q A v

\$\$

where \$n\$ is the number density of charge carriers, \$q\$ the charge of each carrier, \$A\$ the cross‑sectional area, and \$v\$ the drift speed.

The total magnetic force on all carriers in the segment is

\$\$

F = (nqAv)L\,B = I L B

\$\$

The direction of the force follows Fleming’s left‑hand rule (or the right‑hand rule for positive charge flow).

Summary Table

Worked Example

1. Calculate the radius of a proton (mass \$1.67\times10^{-27}\,\text{kg}\$, charge \$+e\$) moving at \$2.0\times10^{6}\,\text{m s}^{-1}\$ in a \$0.5\,\text{T}\$ magnetic field perpendicular to its velocity.

   Using \$r = \dfrac{mv}{qB}\$:

   \$\$

   r = \frac{(1.67\times10^{-27}\,\text{kg})(2.0\times10^{6}\,\text{m s}^{-1})}{(1.60\times10^{-19}\,\text{C})(0.5\,\text{T})}

   \approx 4.2\times10^{-2}\,\text{m}

   \$\$

2. Find the force on a \$0.30\,\$m long wire carrying \$5.0\,\$A placed in the same \$0.5\,\$T field, perpendicular to the wire.

   Using \$F = I L B\$:

   \$\$

   F = (5.0\ \text{A})(0.30\ \text{m})(0.5\ \text{T}) = 0.75\ \text{N}

   \$\$

Key Points to Remember

• The magnetic force on a moving charge is always perpendicular to both the velocity and the magnetic field.
• When \$\mathbf{v}\perp\mathbf{B}\$ the particle undergoes uniform circular motion.
• The radius of curvature \$r\$ is proportional to the particle’s momentum and inversely proportional to its charge and the magnetic field strength.
• For a current‑carrying conductor the total magnetic force is \$F = I L B\sin\theta\$; with \$\theta=90^\circ\$ this reduces to \$F = I L B\$.
• The direction of the force on a conductor is given by Fleming’s left‑hand rule (or the right‑hand rule for positive charge flow).