Magnetism and Magnetic Effects: Speed-Solving Techniques (4)

Question

A wire of length $L = 0.5$ m carries a current $I = 4$ A in a uniform magnetic field $B = 0.3$ T. Find the force on the wire when (a) the wire is perpendicular to $B$, (b) the wire makes an angle $30^\circ$ with $B$, and (c) the wire is parallel to $B$.

Solution — Step by Step

Final answers: (a) $0.6$ N, (b) $0.3$ N, (c) $0$ N.

Why This Works

The magnetic force on a moving charge is $\vec{F} = q\vec{v}\times\vec{B}$. Summing over all charge carriers in a wire of length $L$ carrying current $I$ gives $\vec{F} = I\vec{L}\times\vec{B}$. The cross product means only the component of velocity (or current direction) perpendicular to $\vec{B}$ contributes.

A wire parallel to $B$ has no perpendicular component, hence zero force — this is why solenoid wires aligned with their own axis feel no self-force.

Alternative Method

For case (b), decompose the wire into components: one along $B$ (length $L\cos 30^\circ$, no force) and one perpendicular (length $L\sin 30^\circ$, full force). The perpendicular piece gives $F = BI(L\sin 30^\circ) = 0.3 \times 4 \times 0.25 = 0.3$ N. Same answer, useful when wires bend.

Common Mistake

Confusing the angle. $\theta$ is the angle between current direction and $\vec{B}$, not between current and the perpendicular to $\vec{B}$. So a wire ”$30^\circ$ from the field” gives $\sin 30^\circ = 0.5$, not $\cos 30^\circ$.