Correct Answer - (a) `6.6 xx 10^(-5) T, (b) 0 , 0, 8 xx 10^(-6) Nt` .
`B =4 xx (mu_(0))/(4pi) (I)/r_(1) xx (pi)/(4) + 4 xx (mu_(0))/(4pi) (I)/(r_(2)) xx (pi)/(4)`
`B = (mu_(0)I)/(4r_(1)) + (mu_(0)I)/(4r_(2)) = (mu_(0)I)/(4) (((1)/(r_(1)) + (1)/(r_(2)))o.`
` B =(12.57 xx 10^(-7))/(4) xx 10 xx [(1)/(0.08) + (1)/(0.12)]`
`B =3.16 xx 10^(-6) xx [(25)/(2) + (25)/(3)]`
`oversetrarrB =6.6 xx 10^(-5) jo.`
(b) Force on the wire at center wire be zero because current in wire will be anti parallel to field at centre
(ii) Force on arc`AC` will also zero becuase field due to wire at `AC` will be tangent to current at each point
`F_(CD) = underset(0.08)overset(0.12)intIB_(x) dx = underset(0.08)overset(0.12)intI(mu_(0))/(4pi) (2I)/(x) dx = (2mu_(0)I^(2))/(4pi)`
` underset(0.08)overset(0.12)int(x)/(dx) = (2mu_(0))/(4pi) I^(2) [log_(C) x]_(0.08)^(0.12) = 2 (mu_(0))/(4pi) I^(2)`
`[log _(e) "(0.12)/(0.08)]=2 xx 10^(-7)xx 100 log_(c) 1.5 =8`
`xx 10^(-6) N`
![image](https://learnqa.s3.ap-south-1.amazonaws.com/images/16110713277042064251611071327.png)
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