Two long parallel wires are 30 cm apart carrying currents 10 A and 15 A respectively in the same direction. The force acting over a length of 5 m of the wires is

Explanation

(i) Let us call `vecB` due to (1) and (2) as `vecB_(1)` and `vecB_(2)` respectively.Then at A:`vecB_(1)` is `o.` and `vecB_(2)` is `ox`
`B_(1)=(mu_(0)2l)/(2pia)`and `B_()=(mu_(0)l)/(2pi2a)`
`:. B_(res)=B_(1)-B_(2)=3/4 (mu_(0)l)/(piepsilon)o.`
at `C:vecB_(1)` is `ox` and `vecB_(2)` is `ox`
`:. B_(res)=B_(1)+B_(2)=(mu_(0)2I)/(2pia/2)+(mu_(0)I)/(2pia/2)=(6mu_(0)I)/(2pia)=(3mu_(0)I)/(pia)ox`
at `D:vecB_(1)` is `ox` and `vecB_(2)` is `ox` and both are equal is magnitude
(ii) It is clear from the above solution that `B=0` at point `D`.