(1) The net magnetic field `vecB` at point P is the vector sum of the magnetic fields due to the currents in the two wires.
(2) We can find the magnetic field due to any current by applying the Biot-Savart law to the current. For points near the current in a long straight wire, that law leads to Eq.29-4.
Finding the vectors : In Fig. 29-15a, point P is distance R from both currents `i_(1)` and `i_(2)`. Thus, Eq. 29-4 tells us that at point P those currents produce magnetic fields `vecB_(1)` and `vecB_(2)` with magnitudes
`B_(1)=(mu_(0)i_(1))/(2piR)` and `B_(2)=(mu_(0)i_(2))/(2piR)`
In the right triangle of Fig. 29-15a, note tht the base angles (between sides R and d) are both `45^(@)`. This allows us to write `cos45^(@)=R//d` and replace `R` with `d cos 45^(@)`. Then the field magnitudes `B_(1)` and `B_(2)` become
`B_(1)=(mu_(0)i_(1))/(2pidcos45^(@))` and `B_(2)=(mu_(0)i_(2))/(2pidcos45^(@))`
We want to combine `vecB_(1)` and `vecB_(2)` to find their vector sum, which is the net field `vecB` at P. To find the directions of `vecB_(1)` and `vecB_(2)`, we apply the right-hand rule of Fig. 29-5 to each current in Fig. 29-15a. For wire 1, with current out of the page , we mentally grasp the wire with the right hand, with the thumb pointing out of the page. Then the curled fingers indicate that the field lines run counterlockwise. In, they are directed upward to the left. Recall that the magnetic field at a point near a long, straight current -carrying wire must be directed perpendicular to a radial line between the point
and the current . Thus `vecB_(1)` must be directed upward to the left as drawn in Fig. 29-15b. (Note carefully the perpendicular symbol between vector `B_(1)` and the line connecting point P and wire 1)
Repeating this analysis for the current in wire 2 we find that `vecB_(2)` is directed upward to the right as drawn in Fig. 29-15b.
Adding the vectors : We can now vectorially add `vecB_(1)` to find the net magnetic field `vecB` at point P, either by using a vector -capable calculator or by resolving the vectors into components and then combining the components of `vecB`. However in Fig. 29-15b, there is a third method : Becuase `vecB_(1)` and `vecB_(2)` are perpendicular to each other, they form the legs of a right triangle, with `vecB` as the hypotenuse. So,
`B=sqrt(R_(1)^(2)+R_(2)^(2))=(mu_(0))/(2pid(cos45^(@)))sqrt(i_(1)^(2)+i_(2)^(2))`
`=((4pixx10^(-7)T*m//A)sqrt((15A)^(2)+(32A)^(2)))/((2pi)(5.3xx10^(-2)m)(cos45^(@)))`
`-1.89xx10^(-4)T~~190muT`.
The angle `phi` between the directions of `vecB` and `vecB_(2)` in Fig. 29-15b follows from
`phi=tan^(-1).(B_(1))/(B_(2))`
which, with `B_(1)` and `B_(2)` as given above, yields
`phi=tan^(-1).(i_(1))/(i_(2))=tan^(-1).(15A)/(32A)=25^(@)`
The angle between `vecB` and the x axis shown in Fig. 29-15b is then
`phi+45^(@)=25^(@)+45^(@)=70^(@)`
