Answer:
523.2 x 10^-7 T
Explanation:
distance between the two wires, D = 20 cm
d = D/2 = 10 cm = 0.1 m
Current in first wire, i1 = 25.6 A
Current in second wire, i2 = 5.4 A
The magnetic field due to a straight current carrying conductor is given by
[tex]B=\frac{\mu _{0}}{4\pi }\frac{2i}{r}[/tex]
Magnetic field due to the first wire
[tex]B_{1}=10^{-7}\times \frac{2\times 25.6}{0.1}[/tex]
[tex]B_{1}=512\times 10^{-7}[/tex] T
Magnetic field due to the second wire
[tex]B_{2}=10^{-7}\times \frac{2\times 5.4}{0.1}[/tex]
[tex]B_{2}=108\times 10^{-7}[/tex] T
The two magnetic field are perpendicular to each other
The resultant magnetic field is given by
[tex]B=\sqrt{B_{1}^{2}+B_{2}^{2}}[/tex]
[tex]B=\sqrt{512^{2}+108^{2}}\times 10^{-7}[/tex]
B = 523.2 x 10^-7 T
Based on the current in the two wires and their distance away from each other, the magnitude of the magnetic field at the midway point will be 5.23 x 10⁻⁵ T.
This can be found by the formula:
= (4π x 10⁻⁷ / 2π x Midway distance in meters) x √(Current in top wire² + Current in bottom wire²)
Midway distance:
= 20 cm /2
= 10 cm
= 0.10 meters
Solving gives:
= (4π x 10⁻⁷ / 2π x 0.10) x √(25.6² + 5.4²)
= 5.23 x 10⁻⁵ T
Find out more on magnetic field magnitude at https://brainly.com/question/26257705.
