Question

What is the net force on the square coil?

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

Answer 1

The correct option is D $25\times {10}^{-7}N$ moving towards wire


We use Fleming’s left-hand rule to calculate direction of force. By symmetry, the forces on conductors $AD$ and $BC$ will cancel out each other.

Force on a conductor $=BiL$

${\overrightarrow{F}}_{AB}=\frac{{\mu }_0{i}_1}{2\pi r}\times i_2\times L$

$=\frac{2\times {10}^{-7}\times 2}{2\times {10}^{-2}}\times 1\times 15\times {10}^{-2}$

$=3\times {10}^{-6}Nleftside$

${\overrightarrow{F}}_{CD}=\frac{{\mu }_0{i}_1}{2\pi r}\times i_2\times L$

$=\frac{2\times {10}^{-7}\times 2}{12\times {10}^{-2}}\times 1\times 15\times {10}^{-2}$

$=0.5\times {10}^{-6}Nrightside$

$F_{net}={\overrightarrow{F}}_{AB}-{\overrightarrow{F}}_{CD}=2.5\times {10}^{-6}N$

$=25\times {10}^{-7}N$ towards the wire.

Hence, option (a) is correct.