Answer:
Resultant magnetic field =[tex]3.3\times 10^{-5} T[/tex]
Explanation:
We are given that two long straight, parallel wires separated by a distance 20 cm
Le two wires A and B
[tex]r_1=15cm=0.15 m,r_2=25cm=0.25m[/tex]
Current flowing in wire=[tex]I_A[/tex] =30 A
Current flowing in wire B=[tex]I_B=[/tex]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 [tex]I_A,B_1=\frac{\mu_0}{4\pi}\times\frac{2I_A}{r_1}[/tex]
Magnetic field due to current [tex]I_A,B_1=[tex]10^{-7}\times \frac{2\times 30}{0.15}=4\times 10^{-5}[/tex] T
Magnetic field due to current [tex]I_B,B_2=10^{-7}\times \frac{2\times 40}{0.25}=10^{-7}\times 320[/tex] T
Magnetic field due to current [tex]I_B,B_2=3.2\times 10^{-5}[/tex]
Angle between [tex]B_1,and\;B_2=\phi=90^{\circ}+\theta[/tex]
[tex]sin\theta=\frac{0.15}{0.25}=\frac{3}{5}[/tex]
Resultant magnetic field =[tex]\sqrt{B^2_1+B^2_2-2B_1B_2Cos\phi}[/tex]
Resultant magnetic field =[tex]\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)}[/tex]
Resultant magnetic field =[tex]\sqrt{16\times 10^{-10}+10.24\times 10^{-10}-25.6\times 10^{-10}\times \frac{3}{5}}[/tex]
Resultant magnetic field =[tex]\sqrt{10.88}\times 10^{-5}[/tex]
Resultant magnetic field =[tex]3.3\times 10^{-5} T[/tex]
