determine the direction of the force on a charge moving in a magnetic field

Force on a Current‑Carrying Conductor

Learning Objective

Determine the direction of the magnetic force on a charge moving in a magnetic field and extend this to a current‑carrying conductor.

1. Magnetic Force on a Single Charge

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

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

Key points:

• The force is perpendicular to both \$\mathbf{v}\$ and \$\mathbf{B}\$.
• The magnitude is \$F = qvB\sin\theta\$, where \$\theta\$ is the angle between \$\mathbf{v}\$ and \$\mathbf{B}\$.
• The direction is found using the right‑hand rule for the cross product.

2. Right‑Hand Rule for \$\mathbf{v}\times\mathbf{B}\$

To apply the rule:

1. Point the fingers of your right hand in the direction of \$\mathbf{v}\$ (the motion of a positive charge).
2. Sweep the fingers toward the direction of \$\mathbf{B}\$.
3. Thumb points in the direction of the force \$\mathbf{F}\$ on a positive charge.
4. For a negative charge, the force is opposite to the thumb direction.

3. From a Single Charge to a Current‑Carrying Conductor

A conductor carrying a current \$I\$ contains many charge carriers (usually electrons) moving with drift velocity \$\mathbf{v}_d\$. The total magnetic force on a length \$L\$ of the conductor is the sum of the forces on all carriers:

\$\mathbf{F} = I\,\mathbf{L}\times\mathbf{B}\$

where \$\mathbf{L}\$ is a vector of magnitude \$L\$ directed along the conventional current (positive to negative).

4. Direction of the Force on a Conductor

Use the same right‑hand rule, but replace \$\mathbf{v}\$ with the direction of the current \$\mathbf{I}\$ (or \$\mathbf{L}\$):

1. Point the fingers in the direction of the conventional current (from positive to negative).
2. Rotate the hand so the fingers can sweep toward the magnetic field direction \$\mathbf{B}\$.
3. The thumb then points in the direction of the force on the conductor.

5. Summary Table of Directions

6. Example Problem

Problem: A straight wire of length \$0.30\ \text{m}\$ carries a current of \$5.0\ \text{A}\$ to the north. It is placed in a uniform magnetic field of \$0.20\ \text{T}\$ directed eastwards. Find the magnitude and direction of the magnetic force on the wire.

Solution:

1. Identify vectors:

   • \$\mathbf{L}\$ points north (positive \$y\$).
   • \$\mathbf{B}\$ points east (positive \$x\$).

2. Use \$F = I L B \sin\theta\$. Here \$\theta = 90^\circ\$, so \$\sin\theta = 1\$.
3. Calculate magnitude:

   \$F = (5.0\ \text{A})(0.30\ \text{m})(0.20\ \text{T}) = 0.30\ \text{N}\$

4. Direction: Apply right‑hand rule.

   • Thumb (current) points north.
   • Fingers point east (field).
   • Thumb of the right hand (force) points upward, i.e., out of the page.

7. Common Misconceptions

• Electron motion vs. conventional current: The right‑hand rule uses conventional current direction. If you consider electron drift, the force direction is opposite.
• Angle \$\theta\$: The force is zero when \$\mathbf{v}\$ (or \$\mathbf{I}\$) is parallel or antiparallel to \$\mathbf{B}\$ because \$\sin 0^\circ = 0\$.
• Force on the wire vs. force on individual charges: The net force on the wire is the sum of forces on all carriers, giving the simple \$I\mathbf{L}\times\mathbf{B}\$ expression.

8. Practice Questions

1. A rectangular loop of wire carries a current \$I\$ clockwise when viewed from above. The loop lies in a uniform magnetic field directed into the page. Determine the direction of the net force on each side of the loop.
2. A proton moves with speed \$2.0\times10^6\ \text{m s}^{-1}\$ perpendicular to a magnetic field of \$0.5\ \text{T}\$. Calculate the magnitude of the magnetic force if the proton’s charge is \$+e\$.
3. Explain why a current‑carrying wire placed parallel to a magnetic field experiences no magnetic force.

9. Key Take‑aways

• The magnetic force on a moving charge is given by \$\mathbf{F}=q\mathbf{v}\times\mathbf{B}\$.
• For a current‑carrying conductor, the force is \$\mathbf{F}=I\mathbf{L}\times\mathbf{B}\$.
• Direction is found using the right‑hand rule with conventional current direction.
• The force is maximal when \$\mathbf{v}\$ (or \$\mathbf{I}\$) is perpendicular to \$\mathbf{B}\$ and zero when they are parallel.