Exercises — Magnetic force on current-carrying conductor

WHAT: perpendicular means $\theta =90^{\circ }$, so $\sin \theta =1$ and the formula collapses to $F=BIL$. WHY: at $90^{\circ }$ the whole wire "cuts across" the field, so every carrier feels the maximum sideways shove. $F=BIL=0.20\times 5.0\times 0.40=0.40\mathrm{N}.$ Answer: $0.40\mathrm{N}$.

Vector form: $\vec{F}=I\vec{L}\times \vec{B}$ with magnitude $F=BIL\sin \theta$.

• Maximum at $\theta =90^{\circ }$ (wire $\perp$ field): $F=BIL$.
• Zero at $\theta =0^{\circ }$ or $180^{\circ }$ (wire $\parallel$ or anti-$\parallel$ to field): $F=0$. WHY zero when parallel: $\vec{L}\parallel \vec{B}\Rightarrow \vec{L}\times \vec{B}=\vec{0}$, because the cross product measures how perpendicular two vectors are.

WHAT: the wire is tilted, so keep the full $\sin \theta$. WHY: only the part of the wire across the field is pushed; $\sin 30^{\circ }=0.5$ says exactly half of the "sideways-ness" survives. $F=BIL\sin \theta =0.50\times 3.0\times 0.60\times \sin 30^{\circ }=0.90\times 0.5=0.45\mathrm{N}.$ Answer: $0.45\mathrm{N}$.

WHAT: we know $F$, $I$, $L$, and $\theta =90^{\circ }$; solve $F=BIL$ for $B$. WHY rearrange: the formula ties four numbers together; knowing any three finds the fourth. $B=\frac{F}{IL}=\frac{0.24}{2.0\times 0.30}=\frac{0.24}{0.60}=0.40\mathrm{T}.$ Answer: $B=0.40\mathrm{T}$.

WHAT: set up axes and use the cross product to double-check the hand. Let $\hat{x}=$ east, $\hat{y}=$ north, $\hat{z}=$ up. Current is north: $\vec{L}\propto \hat{y}$. Field is down: $\vec{B}\propto -\hat{z}$. $\vec{L}\times \vec{B}\propto \hat{y}\times (-\hat{z})=-(\hat{y}\times \hat{z})=-\hat{x}=\text{west}.$ Answer: the force points west. Hand check (Fleming's Left-Hand Rule, defined at the top of this page): First finger down (field), second finger north (current) → thumb points west. ✓

WHAT: compute $\vec{F}=I\vec{L}\times \vec{B}$ component-by-component. WHY the cross product: the $\hat{x}$-part of $\vec{B}$ is parallel to the wire and must contribute nothing; the cross product automatically kills it.

WHAT: use the uniform-field shortcut from the parent note. In a uniform $\vec{B}$, $\vec{F}=I(\int d\vec{\ell })\times \vec{B},$ and $\int d\vec{\ell }$ is just the straight vector from start end to finish end — the diameter, length $2R$. WHY this saves us: we avoid integrating force around the curve; only the endpoints matter. $L_{\text{eff}}=2R=0.40\mathrm{m}.$ The diameter is perpendicular to $\vec{B}$ (which is out of the page), so $\theta =90^{\circ }$: $F=BI{L}_{\text{eff}}=0.30\times 6.0\times 0.40=0.72\mathrm{N}.$ Direction — derived, not just read off: put the diameter along the $x$-axis, with the effective current vector pointing from the start end to the finish end, ${\vec{L}}_{\text{eff}}=2R\hat{x}$. The field is out of the page, $\vec{B}=B\hat{z}$. Then $\vec{F}=I{\vec{L}}_{\text{eff}}\times \vec{B}=I(2R)B(\hat{x}\times \hat{z})=I(2R)B(-\hat{y}),$ using $\hat{x}\times \hat{z}=-\hat{y}$. So the force points along $-\hat{y}$: perpendicular to the diameter, in the plane of the page, pointing away from the bulge of the semicircle. The orange arrow in the figure shows this direction. Answer: $0.72\mathrm{N}$, perpendicular to the diameter.

WHAT: apply the same shortcut, but now the wire returns to its start. $\vec{F}=I(\oint d\vec{\ell })\times \vec{B}.$ For any closed path, $\oint d\vec{\ell }=\vec{0}$ (you end where you began, so total displacement is zero). $\vec{F}=I(\vec{0})\times \vec{B}=\vec{0}.$ Answer: net force is zero. Each side still feels a force, but opposite contributions cancel. What survives is a torque — the seed of the current-loop torque and the motor.

WHAT: set the upward magnetic force equal to the downward weight, per unit length. WHY per unit length: both forces scale with $L$, so $L$ cancels and we get a clean condition. Weight per length: $\frac{mg}{L}=\lambda g$. Magnetic force per length (at $\theta =90^{\circ }$): $\frac{F}{L}=BI$. Balance: $BI=\lambda g\Rightarrow I=\frac{\lambda g}{B}=\frac{0.050\times 9.8}{0.40}.$ $I=\frac{0.49}{0.40}=1.225\mathrm{A}\approx 1.2\mathrm{A}.$ Answer: $I\approx 1.2\mathrm{A}$, flowing in the direction that (by Fleming's Left-Hand Rule, defined at the top of this page) makes the force point up.

WHAT: add the fields first (fields superpose), then take one cross product. WHY superpose: magnetic fields add as vectors, so the wire only ever "feels" the total ${\vec{B}}_{\text{tot}}$.

WHAT: one vertical side of one turn is a straight wire of length $\ell$ perpendicular to $\vec{B}$, so it feels $F_1=BI\ell$. Because $N$ turns are stacked at that side, all carrying the same current the same way, their forces add straight up. WHY multiply by $N$: the $N$ wires lie side by side in the same field pointing the same way, so their equal forces simply sum — that is exactly why more turns make a galvanometer more sensitive. $F=NBI\ell =50\times 0.10\times (2.0\times 10^{-3})\times \mathrm{0.030.}$ Step by step: $50\times 0.10=5.0$; then $5.0\times 2.0\times 10^{-3}=1.0\times 10^{-2}$; then $1.0\times 10^{-2}\times 0.030=3.0\times 10^{-4}$. $F=3.0\times 10^{-4}\mathrm{N}.$ Answer: $F=3.0\times 10^{-4}\mathrm{N}$ on one vertical side. WHY the radial field matters: because it keeps $\theta =90^{\circ }$ at every deflection angle, $F$ (and hence the torque) stays proportional to $I$ — that is what makes the galvanometer's scale linear.

WHAT/WHY: The two opposite sides of the loop carry current in opposite directions, so by $\vec{F}=I\vec{L}\times \vec{B}$ they feel equal and opposite forces — the vector sum ($\oint d\vec{\ell }=0$) is zero, so the loop is not pushed anywhere as a whole. But those two equal-opposite forces act along different lines (they are offset by the width of the loop), forming a couple. A couple produces a torque with no net force — it rotates the loop. That rotation, with a commutator to flip the current each half-turn, is the electric motor.

Perpendicular force, $L=0.40$, $I=5.0$, $B=0.20$? ::: $0.40\mathrm{N}$. Force at $30^{\circ }$, $L=0.60$, $I=3.0$, $B=0.50$? ::: $0.45\mathrm{N}$. Field from $F=0.24$, $I=2.0$, $L=0.30$, $90^{\circ }$? ::: $0.40\mathrm{T}$. Semicircle net force ($R=0.20$, $I=6.0$, $B=0.30$)? ::: $0.72\mathrm{N}$ (use $L_{\text{eff}}=2R$). Floating-wire current ($\lambda =0.050$, $B=0.40$, $g=9.8$)? ::: $\approx 1.2\mathrm{A}$. Galvanometer side force ($N=50$, $B=0.10$, $I=2.0\mathrm{mA}$, $\ell =0.030$)? ::: $3.0\times 10^{-4}\mathrm{N}$. Net force on any closed loop in uniform $B$? ::: Zero; torque may remain.