Question

The resultant force on the current loop $PQRS$ due to a long current carrying conductor will be

• A. $10^{-4}N$
• B. $3.6\times 10^{-4}N$
• C. $1.8\times 10^{-4}$
• D. $5\times 10^{-4}N$

Answer 1

Answer: (D)

Force on $SR$ and $PQ$ are equal but opposite so their net will be zero.
Force between two parallel conductors carrying currents $I_1$ and $I_2$
$F=\frac{{\mu }_0}{2\pi }\frac{I_1{I}_{2}l}{r}$
where $r=$ distance between two parallel conductors
$F_{PS}=\frac{10^{-7}\times 2\times 20\times 20\times 15\times 10^{-2}}{2\times 10^{-2}}$
$=6\times 10^{-4}N$
$F_{QR}=\frac{10^{-7}\times 2\times 20\times 20\times 15\times 10^{-2}}{12\times 10^{-2}}$
$=1\times 10^{-4}N$
$F_{\text{net }}=F_{PS}-F_{QR}$
$=6\times 10^{-4}-1\times 10^{-4}$
$=5\times 10^{-4}N$