Question

What is the net force on the square coil?

• A. $25\times 10^{-7}N$ moving towards wire
• B. $25\times 10^{-7}N$ moving away from wire
• C. $35\times 10^{-7}N$ moving towards wire
• D. $35\times 10^{-7}N$ moving away from wire

Answer 1

Answer: (A)

Force on side $BC$ and $AD$ are equal but opposite so their net will be zero.

But $F_{AB}=10^{-7}\times \frac{2\times 2\times 1}{2\times 10^{-2}}\times 15\times 10^{-2}$
$=3\times 10^{-6}N$
and $F_{CD}=10^{-7}\times \frac{2\times 2\times 1}{\left(12\times 10^{-2}\right)}\times 15\times 10^{-2}$
$=0.5\times 10^{-6}N$
$\Rightarrow F_{\text{net }}=F_{AB}-F_{CD}=2.5\times 10^{-6}N$
$=25\times 10^{-7}N$, towards the wire.