Source: High school physics (Chinese)
Problem
Each diagram in Figure shows a current-carrying wire placed in a magnetic field, with two of the three quantities (current $I$, magnetic flux density $B$, and Ampere force $F$) indicated. Determine the direction of the third quantity in each case.
- Left diagram: $F$ points upward and $B$ points to the right. Find the direction of $I$.
- Middle diagram: $I$ points into the page ($\otimes$) and $F$ points to the right. Find the direction of $B$.
- Right diagram: $I$ points into the page ($\otimes$) and $B$ points to the right. Find the direction of $F$.
(1) $I$ is directed out of the page ($\odot$).
(2) $B$ is directed upward.
(3) $F$ is directed downward.
Apply the relation $\vec{F} = I\vec{L}\times\vec{B}$ (or equivalently the left-hand rule). The three vectors $\vec{I}$, $\vec{B}$, $\vec{F}$ are mutually perpendicular and form a right-handed triad.
(1) With $\vec{F}$ up and $\vec{B}$ right, $\vec{I}$ must point out of the page so that $\vec{I}\times\vec{B}$ gives an upward force.
(2) With $\vec{I}$ into the page and $\vec{F}$ to the right, $\vec{B}$ must point upward, since $(-\hat{z})\times\hat{y} = \hat{x}$.
(3) With $\vec{I}$ into the page and $\vec{B}$ to the right, $\vec{F}=I\vec{L}\times\vec{B}$ points downward, since $(-\hat{z})\times\hat{x} = -\hat{y}$.