Question

The resultant force on the current loop $PQRS$ due to a long current carrying conductor will be, if the current flow in the loop is clockwise,

• A. $10^{- 4} \, N$
• B. $3.6\times 10^{- 4} \, N$
• C. $1.8\times 10^{- 4} \, N$
• 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
$F_{P S}=\frac{2 \times 10^{- 7} \times 30 \times 20 \times 10 \times 10^{- 2}}{10^{- 2}}$
$ \, \, =6\times 10^{- 4}$
$F_{Q R}=\frac{2 \times 10^{- 7} \times 30 \times 20 \times 10 \times 10^{- 2}}{12 \times 10^{- 2}}$
$ \, \, =10^{- 4}$ N
$F_{n e t}=F_{P S}-F_{Q R}$
$ \, \, =6\times 10^{- 4}$
$\text{ }=5\times 10^{- 4}N$