Correct Answer - `z =0, x = +- (d)/sqrt3, (ii) (I)/(2pid)sqrt((mu_(0))/(pilambda))`
Let field at `P` is zero
(i) `B_9p0 =0 implies oversetrarr(B_(1)) + oversetrarr(B_(2)) + oversetrarrB_(3) = 0 `
`(mu_(0))/(4pi) (2i)/((x +x)) ox + (mu_(0))/(4pi) (2i)/(x) ox + (mu_(0))/(4pi) (2i)/((d -x)) = 0`
`(1)/(d +x) + (1)/(x) - (1)/(d -x) = 0 implies x =+-(d)/sqrt3`
(ii) When wire (2) is displaced by x along y-axis force is taken per unit lenght
`oversetrarr(F_(2)) = oversetrarr(F_(12)) + oversetrarr(F_(23))`
since `F_(12) = F_(23) = (mu_(0))/(4pi)(2i^(2))/(sqrt(d^(2) + x^(2)))`
`F_(2) =2F_(12) sintheta`
`F=2 (mu_(0))/(4pi) (2i^(2))/(sqrt(d^(2) +x^(2))) xx (x)/(sqrt(d^(2) + x^(2))) = (mu_(0))/(pi) (i^(2)x)/((d^(2) + x^(2)))`
`x ltltd`
`F_(R) = -(mu_(0))/(pi) (i^(2)x)/(d^(2))`
comparing with `F_(R) = - Kx implies K = (mu_(0))/(pi)(i^(2)X^(2))/(d^(2))`
` n = (1)/(2pi) sqrt((K)/(m)) = (1)/(2pi) sqrt((mu_(0)i^(2))/(pid^(2) xx lamda)) = (i)/(2pid) sqrt(mu_(0)/(pilambda))`
![image](https://learnqa.s3.ap-south-1.amazonaws.com/images/16110712001127194761611071200.png)
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