Two long, straight wires a and b are `2*0` m apart,
perpendicular to the plane of the paper as shown in figure
. The wire a carries a current of `9*6 A` directed
into the plane of the figure,The magnetic field at the
point P at a distance of 10/11 m from the wire b is zero.
Find (a) the magnitude and direction of the current in b,
(b) the magnitude of the magnetic field B at the point s
and (c) the force per unit length on the wire b`.

Explanation

Solution:(a) For the magnetic field at p to be zero, the current
in the wire b should be coming out of the plane of the
figure so that the fields due to a and b may be opposite
at p. The magnitudes of these fields should be equal,
so that
` ((mu_0) (9.6A)/ 2pi ( 2+(10/11))m) = (m_0)i/ (2 pi (10/11)m)`
` or, i=3*0 A.`
` (b) ((ab)^2= 4m^2) `
`((as)^2 = 2*56m^2`)
` and ((bs)^2= 1*44m ^2)`
` so that ((ab)^2)=((as)^2) + ((bs)^2) and angle asb=90^@.`
` The field at s due to the wire a `
`= ((mu_0)(9*6 A)/(2 pi xx 1*6 m ))= ((mu_0)/2pi) xx 6 AM^-1 `
and that due to the wire b
` = ((mu_0)/(2pi)) (3A/ 1.2m) = ((mu_0)/(2pi)) xx 2.5 Am^-1.`
` These fields are at 90^@ to each other so that their
resultant will have a magnitude
` ( sqrt((((mu-0)/(2pi)) xx 6Am^-1)^2)+ ((((mu_0)/2pi) xx 2*5 Am^-1)^2))`
` = (mu_0/ 2pi) ( sqrt (36+6*25) Am^-1)`
` = 2 xx (10^-7) (TmA^-1) xx 6.5Am^-1`
= 1.3 xx (10^-6)T.`
` (c) The force per unit length on the wire b
`=((mu_0)(i_1)(i_2))/2pi d ) = (2 xx (10^-7) TmA^-1 xx 6*5 Am^-1) xx ((9*6 A) (3 A)/ (2*0 m))`
` = 2*9 xx (10^-6) N`.