Question

What is the net force on the 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

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


Force on sides BC and CD cancel each other.
Force on side AB $F_{AB}={10}^{-7}\times \frac{2\times 2\times 1}{2\times {10}^{-2}}\times 15\times {10}^{-2}=3\times {10}^{-6}N$
Force on side CD $F_{CD}={10}^{-7}\times \frac{2\times 2\times 1}{12\times {10}^{-2}}\times 15\times {10}^{-2}=0.5\times {10}^{-6}N$
Hence net force on loop = $F_{AB}-F_{CD}=25\times {10}^{-7}N$ (towards the wire).