Magnetic Fields: Question 2

Worked solution

Part (a): Finding the magnetic flux density

The wire is perpendicular to the field, so $\theta=90°$ and $\sin\theta=1$. Rearranging $F=BIL\sin\theta$ for $B$: $B=\frac{F}{IL\sin\theta}=\frac{0.60}{5.0\times0.40\times\sin90°}=\frac{0.60}{2.0}=0.30\text{ T}$

Recomputing as a check: $5.0\times0.40=2.0$, and $0.60\div2.0=0.30\text{ T}$, consistent.

Part (b): Doubling the current

$B$, $L$ and $\theta$ (the wire is still perpendicular, so $\theta=90°$) are all unchanged, so $F=BIL\sin\theta$ is directly proportional to $I$ alone. Doubling $I$ from $5.0\text{ A}$ to $10.0\text{ A}$ therefore doubles the force: $F'=B\times(2I)\times L\times\sin90°=2\times(BIL\sin90°)=2\times0.60=1.2\text{ N}$

Recomputing directly from the formula as a check: $F'=0.30\times10.0\times0.40\times1=1.2\text{ N}$, the same answer both ways.

Part (c): Wire at 30° to the field

The current returns to $5.0\text{ A}$, $B=0.30\text{ T}$ (from part (a)) stays the same, but now $\theta=30°$ instead of $90°$: $F''=BIL\sin30°=0.30\times5.0\times0.40\times0.5=0.30\text{ N}$

Recomputing as a check: $0.30\times5.0=1.5$, then $1.5\times0.40=0.60$, then $0.60\times0.5=0.30\text{ N}$, the same result.

As a further sanity check, since $B$, $I$ and $L$ are the same as in the original $90°$ case, the new force must simply scale by $\sin30°/\sin90°=0.5$: $0.60\times0.5=0.30\text{ N}$, exactly half the original force, as expected, since $30°$ gives less than the maximum ($90°$) force.

Final answers

• (a) $B=\boxed{0.30}\text{ T}$
• (b) The force doubles to $F'=\boxed{1.2}\text{ N}$, because $F\propto I$ with $B$, $L$ and $\theta$ unchanged
• (c) $F''=\boxed{0.30}\text{ N}$