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

Resultant magnetic field =
`3.3* 10^(-5) T`

Step-by-step 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=(\mu_0)/(4\pi)*(2I_A)/(r_1)`

Magnetic field due to current
`I_A,B_1=[tex]10^(-7)* (2* 30)/(0.15)=4* 10^(-5)` T

Magnetic field due to current
`I_B,B_2=10^(-7)* (2* 40)/(0.25)=10^(-7)* 320` T

Magnetic field due to current
`I_B,B_2=3.2* 10^(-5)`

Angle between
`B_1,and\;B_2=\phi=90^(\circ)+\theta`

`sin\theta=(0.15)/(0.25)=(3)/(5)`

Resultant magnetic field =
`√(B^2_1+B^2_2-2B_1B_2Cos\phi)`

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

Resultant magnetic field =
`\sqrt{16* 10^(-10)+10.24* 10^(-10)-25.6* 10^(-10)* (3)/(5)}`

Resultant magnetic field =
`√(10.88)* 10^(-5)`

Resultant magnetic field =
`3.3* 10^(-5) T`