Force on Current-Carrying Conductor

Macroscopic Lorentz force on a straight conductor of length L carrying current I in field B. Derived by summing qv × B over all free charges in the volume.

From Microscopic to Macroscopic

Consider a conductor of length $L$ and cross-sectional area $A$ carrying current $I$ in field $\vec{B}$.

Each free electron drifts with velocity $\vec{v}_d$. The force on one electron: $q\vec{v}_d \times \vec{B}$.

Number of free electrons in the segment: $N = nAL$, where $n$ is carrier density.

Total force:

$$\vec{F} = Nq(\vec{v}_d \times \vec{B}) = nALq(\vec{v}_d \times \vec{B})$$

Since $I = nqAv_d$ and the direction of conventional current $\hat{l}$ is opposite to electron drift:

$$\boxed{\vec{F} = I(\vec{L} \times \vec{B})}$$

where $\vec{L}$ is along the direction of conventional current with magnitude $L$.

Magnitude

$$F = BIL\sin\theta$$

$\theta$ is the angle between the current direction and $\vec{B}$.

• Maximum force when conductor is perpendicular to $\vec{B}$.
• Zero force when conductor is parallel to $\vec{B}$.

For Curved Conductors

For a curved segment, integrate over elements:

$$\vec{F} = I\int d\vec{l} \times \vec{B}$$