Question

Two long, straight, parallel wires separated by a distance of 20 cm carry currents of 30 A and 40 A in opposite directions. What is the magnitude of the resulting magnetic field at a point that is 15 cm from the wire carrying the 30-A current and 25 cm from the other wire?

Answer 1

Answer:

Resultant magnetic field =$3.3\times 10^{-5} T$

Explanation:

We are given that two long straight, parallel wires separated by a distance 20 cm

Le two wires A and B

$r_1=15cm=0.15 m,r_2=25cm=0.25m$

Current flowing in wire=$I_A$ =30 A

Current flowing in wire B=$I_B=$40 A

We have to find the magnitude of magnetic field at a point 15 cm from wire A and 25 cm  from wire B

Magnetic field due to current $I_A,B_1=\frac{\mu_0}{4\pi}\times\frac{2I_A}{r_1}$

Magnetic field due to current $I_A,B_1=[tex]10^{-7}\times \frac{2\times 30}{0.15}=4\times 10^{-5}$ T

Magnetic field due to current $I_B,B_2=10^{-7}\times \frac{2\times 40}{0.25}=10^{-7}\times 320$ T

Magnetic field due to current $I_B,B_2=3.2\times 10^{-5}$

Angle between $B_1,and\;B_2=\phi=90^{\circ}+\theta$

$sin\theta=\frac{0.15}{0.25}=\frac{3}{5}$

Resultant magnetic field =$\sqrt{B^2_1+B^2_2-2B_1B_2Cos\phi}$

Resultant magnetic field =$\sqrt{(4\times 10^{-5})^2+(3.2\times 10^{-5})^2-2\times 4\times 10^{-5}\times 3.2\times 10^{-5} cos (90^{\circ}+\theta)}$

Resultant magnetic field =$\sqrt{16\times 10^{-10}+10.24\times 10^{-10}-25.6\times 10^{-10}\times \frac{3}{5}}$

Resultant magnetic field =$\sqrt{10.88}\times 10^{-5}$

Resultant magnetic field =$3.3\times 10^{-5} T$