A) 9 cm
B) 7 cm
C) 5 cm
D) 3 cm
Correct Answer: A
Solution :
For no force on wire C, force on wire C due to wire D= force on wire C due to wire B \[\Rightarrow \frac{{{\mu }_{0}}}{4\pi }\times \frac{2\times 15\times 5}{x}\times l=\frac{{{\mu }_{0}}}{4\pi }\times \frac{2\times 5\times 10}{\left( 15-x \right)}\times l\]\[\Rightarrow x=9cm.\]You need to login to perform this action.
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