Question

Two parallel long, straight wires are 0.0040 m apart and carry currents of 5.0 A and 10 A, respectively. Determine the net magnetic field midway between the wires. Take B as positive if it is going into the page.Group of answer choices-1.0 x 10-3 T5.0 x 10-4 T-5.0 x 10-4 T1.0 x 10-3 T

Answer 1

Given

Current in the wires,

`\begin{gathered} I_1=5.0\text{ A} \\ I_2=10\text{ A} \end{gathered}`

The distance between the wires,

`d=0.0040\text{ m}`

To find:

The magnetic field between the wires.

Step-by-step explanation:

Let O be the mid point between the wires.

The magnetic field due to the wire carrying the current

`I_1=5\text{ A}`
`\begin{gathered} B_1=(\mu_oI_1)/(2\pi((d)/(2))) \\ \Rightarrow B_1=(4\pi*10^(-7)*5)/(2\pi((0.0040)/(2))) \\ \Rightarrow B_1=2*10^(-7)(5)/(0.002) \\ \Rightarrow B_1=0.0005\text{ T} \end{gathered}`

The magnetic field goes into the plane of the paper.

Again,

The magnetic field due to the wire carrying the current

`I_2=10\text{ A}`
`\begin{gathered} B_2=(\mu_oI_2)/(2\pi((d)/(2))) \\ \Rightarrow B_2=(4\pi*10^(-7)*10)/(2\pi*0.002) \\ \Rightarrow B_2=0.001\text{ T} \end{gathered}`

The magnetic fields goes out of the plane of the paper.

Thus the resultant magnetic field is:

`\begin{gathered} B=B_1-B_2 \\ \Rightarrow B=0.0005-0.001 \\ \Rightarrow B=0.0005\text{ T} \\ \Rightarrow B=-5*10^(-4)T \end{gathered}`

Conclusion

Thus the required answer is:

`-5*10^(-4)T`