Question

The resultant force on the square current loop PQRS due to a long current-carrying conductor will be:
(current flow in the loop is clockwise)

• A. zero
• B. $0.36\times {10}^{-3}N$
• C. $1.8\times {10}^{-3}N$
• D. $0.5\times {10}^{-3}N$

Answer 1

Answer: (D)

$0.5\times {10}^{-3}N$

Since current in the lengths PQ and RS are equal and opposite, the forces also will be equal and opposite. Hence both will cancel each other.
Force on PS :
$F_{PS}=\frac{{\mu }_{0}a{I}_1{I}_2}{2\pi d}$ where $I_1$ and $I_2$ are current in the infinite wire and SP respectively, a is the side length of the square loo and d is the distance of SP from the infinite wire.
$\therefore \overrightarrow{F_{SP}}=\frac{4\pi \times {10}^{-7}\times 10\times {10}^{-2}\times 30\times 20}{2\pi \times 2\times {10}^{-2}}=0.6\times {10}^{-3}\left(\hat{i}\right)$
$\therefore \overrightarrow{F_{QR}}=\frac{4\pi \times {10}^{-7}\times 10\times {10}^{-2}\times 30\times 20}{2\pi \times 12\times {10}^{-2}}=0.1\times {10}^{-3}(-\hat{i})$
Net force on square loop $\overrightarrow{F_{SP}}+\overrightarrow{F_{QR}}=0.5\times {10}^{-3}N$