Question

Two long wires are oriented so that they are perpendicular to each other, and at their closest they are 20.0 cm apart. What is the magnitude of the magnetic field at a point midway between them if the top one carries a curent of 25.6 A and the bottom one carries 5.4 A?

Answer 1

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

$B=\frac{\mu _{0}}{4\pi }\frac{2i}{r}$

Magnetic field due to the first wire

$B_{1}=10^{-7}\times \frac{2\times 25.6}{0.1}$

$B_{1}=512\times 10^{-7}$ T

Magnetic field due to the second wire

$B_{2}=10^{-7}\times \frac{2\times 5.4}{0.1}$

$B_{2}=108\times 10^{-7}$ T

The two magnetic field are perpendicular to each other

The resultant magnetic field is given by

$B=\sqrt{B_{1}^{2}+B_{2}^{2}}$

$B=\sqrt{512^{2}+108^{2}}\times 10^{-7}$

B = 523.2 x 10^-7 T