Magnetic Fields: Question 8

Worked solution

Part (a): Force on a single turn

The side is at right angles to the field, so $\theta=90°$ and $\sin\theta=1$. Using $F=BIL\sin\theta$ for one turn: $F_1=0.25\times1.5\times0.12\times\sin90°=0.045\text{ N}$

Recomputing as a check: $0.25\times1.5=0.375$, and $0.375\times0.12=0.045\text{ N}$, the same result.

Part (b): Total force on all 60 turns

Each of the $N=60$ turns carries the same current in the same field, so the total force is $N$ times the force on one turn: $F_{\text{total}}=N\times F_1=60\times0.045=2.7\text{ N}$

Recomputing directly from $F_{\text{total}}=NBIL\sin\theta$ as a check: $60\times0.25=15$, then $15\times1.5=22.5$, then $22.5\times0.12=2.7\text{ N}$, the same result both ways.

Part (c): Coil rotated to 40° to the field

$N$, $B$, $I$ and $L$ are all unchanged, but now $\theta=40°$ instead of $90°$: $F_{\text{total}}'=NBIL\sin40°=2.7\times\sin40°$

Using $\sin40°=0.6428$: $F_{\text{total}}'=2.7\times0.6428=1.736\text{ N}$

Recomputing directly from the full formula as a check: $60\times0.25\times1.5\times0.12=2.7$ (the $90°$ value from part (b)), and $2.7\times0.6428=1.736\text{ N}$, consistent. As a further sanity check, since $\sin40°<\sin90°$, the force at $40°$ must be smaller than at $90°$, which it is ($1.7\text{ N}<2.7\text{ N}$), as expected.

Final answers

• (a) $F_1=\boxed{0.045}\text{ N}$
• (b) $F_{\text{total}}=\boxed{2.7}\text{ N}$
• (c) $F_{\text{total}}'=\boxed{1.7}\text{ N}$ (to 2 s.f.)