Two long straight parallel wires carry currents of 8 A and 10 A in opposite directions. If the distance of separation between the wires is 9 cm , then the net magnetic field at a point between the two wires, which is at a perpendicular distance of 4 cm from the wire carrying 8 A current is
Step 1: Find the magnetic field due to the first wire (8A) The formula for the magnetic field at a distance r from a straight wire carrying current I is: B=‌
µ0I
2Ï€r
For the first wire: I1=8A,r1=4cm=4×10−2m So, B1=‌
µ0×8
2π∨4∨10−2
(−
∧
k
)‌ (The ‌−
∧
k
‌ shows the direction of the field.) ‌
Step 2: Find the magnetic field due to the second wire (10 A) The second wire has I2=10A and is r2=5cm=5×10−2m away from the point. (Since the wires are 9 cm apart and the point is 4 cm from the first wire, it must be 5 cm from the second wire.) So, B‌II ‌=‌
µ0×10
2π×5×10−2
(−
∧
k
) Step 3: Add the two magnetic fields Both fields point in the same direction (−
∧
k
), so add their values: Bnet=B1+BII=‌
µ0
2π×10−2
[‌
8
4
+‌
10
5
](−
∧
k
) Calculate inside the brackets: ‌
8
4
=2,‌
10
5
=2 So, B‌net ‌=‌
µ0
2π×10−2
[2+2](−
∧
k
)=‌
µ0
2π×10−2
⋅4(−
∧
k
) Step 4: Substitute the value of µ0 We know µ0=4π×10−7Tm∕A. So, B‌net ‌=‌