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. $1.8\times10^{-4}\,$
• D. $5\times10^{-4}\, N$

Answer 1

Answer: (D)

Force on $S R$ and $P Q$ 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_{P S} =\frac{10^{-7} \times 2 \times 20 \times 20 \times 15 \times 10^{-2}}{2 \times 10^{-2}} $
$=6 \times 10^{-4} N$
$F_{Q R} =\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_{P S}-F_{Q R} $
$=6 \times 10^{-4}-1 \times 10^{-4} $
$=5 \times 10^{-4}\, N$