Current `I_1 and I_2` flow in the wires shown in Fig. The field is zero at distance x to the right of O. Then

Options

A. `x=((I_(1))/(I_(2)))a`
B. `x=((I_(2))/(I_(1)))a`
C. `x=((I_(1)-I_(2))/(I_(1)+I_(2)))a`
D. `x=((I_(1)+I_(2))/(I_(1)-I_(2)))a`

Answer

C

Explanation

`(mu_(0)I_(1))/(4pi(a+x))=(mu_(0)I_(2))/(4pi(a-x))`
`(a-x)/(a+x)=(I_(2))/(I_(1))`
`I_(1)a-I_(1)x=I_(2)a-I_(2)x`
`x=((I_(1)-I_(2))/(I_(1)+I_(2)))a`