Question

Please help me complete. The second picture is the options you have to choose from from each

Answer 1

This drawing helps us understand the contribution of each the magnetic field of each wire. Wire 1 will contribute with positive field for everything on its left, and with negative field for everything on its right. The opposite is true for wire 2, so, if we analyse regions I, II and III:

Region I: Wire 1 contributes with positive and Wire 2 contributes with negative. As any point on region I will be closer to Wire 1, its contribution will be more significant, and the field will be positive

Region II: Wire 1 and 2 contribute with negative field. Thus, the resulting field will be negative

Region III: Wire 1 contributes with negative field and Wire 2 contributes with positive field. As any point on region III will be closer to Wire 2, its contribution will be more significant, and the field will be positive.

Now all we have to do is calculate the value of the fields. The field on point P1 will have positive contribution from Wire 2 and negative contribution from Wire 1, which will be:

`P1=(\mu_0*I)/(2\pi d)-(\mu_0*I)/(2\pi(2d))=(4\pi *10^(-7))/(2\pi *11.4*10^(-2))-(4\pi *10^(-7))/(4\pi *11.4*10^(-2))=0.8772\mu T`

The field on the midpoint (distance d/2 from each wire) will be negative, as the contribution from both wires will be negative, thus:

`P_(med)=-2*(4\pi *10^(-7)*5.4)/(2\pi *((11.4)/(2))*10^(-2))=-37.8947\mu T`

The field on point P2 will be positive, which can be written as:

`P2=(\mu_0I)/(2\pi *2d)-(\mu_0I)/(2\pi *(3d))=(4\pi *10^(-7)*5.4)/(2\pi *(2*11.4*10^(-2)))-(4\pi *10^(-7)*5.4)/(2\pi *(3*11.4*10^(-2)))`
`P2=1.5789\mu T`

Summary:

a) Pmed = -37.8947uT into the screen

b) P1 = 0.8772uT out of the screen

c) P2 = 1.5789uT out of the screen