Figure shows two parallel wires separated by a distance of 4.0 cm and carrying equal currents of 10 A along opposite directions. Find the magnitude of the magnetic field B at the points A1, A2, A3 and A4.


Given:
Magnitude of current in both the wires: i = 10 A
Distance between 2 wires : d = 4cm = 0.04 m

P and Q are the two wires having opposite direction of currents. From the diagram, P has current into the plane and Q has current coming out of the plane.
By right hand rule, P will have field coming up at A1 and going down at A2 and A3 and tangent to A4 as shown by the red arrow.
Similarly, Q will have field going down as A1,A2 and A3 aand tangent to A4 as shown by blue arrow.
Distance between, A1,A2 and A3 is 2 cm each.
Formula used:

By Ampere’s Law for a current carrying wire is
Here, B is the magnitude of magnetic field, μ0 is the permeability of free space and μ0= 4π × 10-7 T mA-1 and d is the distance between the current carrying wire and the required point.
At A1 :
Net magnetic field due to P and Q would be subtraction of individual fields at A1.
PA1 :d1 = 0.02 m
QA1 : d2 = 0.06m
Bnet = BP – BQ
Where, BP is the magnetic field due to wire P and BQ is the magnetic field due to wire Q.


∴ Bnet = 6.66 × 10-5 T
Hence, magnetic field at A1 is 6.66 × 10-5 T.
At A2
Effect of magnetic field at A2 will be sum of magnetic field due to P and Q as both are in same direction as shown.
PA2: d1 = 0.01 m
QA3: d2 = 0.03 m
Bnet = BP + BQ
Where, BP is the magnetic field due to wire P and BQ is the magnetic field due to wire Q.



∴ Bnet = 2.66 × 10-4 T
Hence , magnetic field at A2 is 2.66 × 10-4 T
At A3
Effect of magnetic field at A3 will be sum of magnetic field due to P and Q as both are in same direction as shown.
PA3: d1 = 0.02 m
QA3 : d2 = 0.02 m
Bnet = BP + BQ
Where, BP is the magnetic field due to wire P and BQ is the magnetic field due to wire Q.



∴ Bnet = 2× 10-4 T
Hence, magnetic field at A3 is 2× 10-4 T
At A4:
The resultant magnetic field at A4 would be
 From Pythagoras theorem,
(PA4)2 = (PA3)2 + (A3A4)2
∴ (PA4)2 = 0.022 + 0.022
∴ PA4 = 0.028 m = d1
Now,

∴BP = 7.14 × 10-5 T
Similarly,
QA4 = 0.028 m = d2
Thus,


∴ BQ = 7.14 × 10-5 T
Substituting to get net magnetic field at A4,

∴ Bnet = 1× 10-4 T
Hence magnetic field at A4 is 1× 10-4 T.


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