A current is a stream of moving charges (Electric Current), and each moving charge feels a magnetic force (Magnetic Force on a Moving Charge). Add up all those small pushes in a wire and you get a force on the wire itself. Every electric motor and loudspeaker runs on it.
Why it works
Take a straight piece of wire of length $L$ carrying current $I$ across a field $B$. If the charges drift at speed $v_d$, it takes them a time $t = L/v_d$ to cross the piece, so the moving charge inside it is $Q = It = IL/v_d$. Each bit feels $qv_dB$, so the total is $$F = Qv_dB = \frac{IL}{v_d}\,v_dB = ILB.$$ The drift speed cancels. At an angle, only the part across the field counts, which brings back $\sin\theta$.
Key idea
A straight wire of length $L$ carrying current $I$ at angle $\theta$ to a uniform field feels $$F = ILB\sin\theta,$$ at right angles to both the wire and the field. Point your fingers along the conventional current and curl them toward $\vec{B}$; your thumb gives $\vec{F}$.
Common mistake
Using the direction the electrons move. In a metal wire the electrons drift opposite to the current, and their charge is negative. Those two flips cancel, so the force is the same either way, but only if you treat it consistently. The rule of this course: use the conventional current $I$ with the right-hand rule and no extra flip. Check: a reversed current must give a reversed force.
Worked example: Force on a power line
A $2.0\ \text{m}$ piece of wire carries $5.0\ \text{A}$ straight across a $0.30\ \text{T}$ field. How big is the force?
- Across the field means $\theta = 90^\circ$, so $\sin\theta = 1$.
- $$F = ILB = (5.0\ \text{A})(2.0\ \text{m})(0.30\ \text{T}) = 3.0\ \text{N}.$$
- Units: $\text{A·m·T} = \text{A·m}\cdot\frac{\text{N}}{\text{A·m}} = \text{N}$.
Answer: $3.0\ \text{N}$
Worked example: Holding a wire up
A horizontal wire has a mass of $0.050\ \text{kg}$ per meter. It runs east–west, at right angles to a horizontal $0.40\ \text{T}$ field. What current makes the magnetic force hold it up?
- For a length $L$, the weight is $mg = (0.050L)(9.8) = 0.49L$ newtons.
- The magnetic force must match it: $ILB = 0.49L$, so the $L$ cancels.
- $I = \dfrac{0.49}{0.40} = 1.2\ \text{A}$, flowing in whichever direction makes $\vec{F}$ point up.
Answer: $I \approx 1.2\ \text{A}$