Question

Two long straight wires enter a room through a window. One carries a current of 3.0 ???? into the room while the other carries a current of 5.0 ???? out. The magnitude in Tm of the path integral ∮ ????⃗ ∙ ???????? around the window frame is:

Answer 1

Answer:

$\begin{equation}\\\oint_LB.dl\\\end{equation}$ = -8πx$10^{-7}$

Explanation:

If you need calculate

$\begin{equation}\\\oint_L B.dl\\ \end{equation}$

You can use the Ampere's Law

$\begin{equation}\\\oint_L B.dl\\ \end{equation}$ = $I_{in}$μ

   

      Where $I_{in}$: Current passing through the window

                   μ : Free space’s magnetic permeability

                   μ = 4πx$10^{-7} T.m.A^{-1}$

Then

$\begin{equation}\\\oint_L B.dl\\ \end{equation}$ = (3-5)4πx$10^{-7}$

$\begin{equation}\\\oint_L B.dl\\ \end{equation}$ = -8πx$10^{-7}$

Answer 2

The magnitude of in the path integral is 2.5*10^-6 T.M

Data;

I1 = 5A

I2 = 3A

μo = 4π * 10^-7 T.m.A^-1

Ampere Law

Ampere law states that the sum of the length of elements multiplied by the magnetic field in the path of the length elements is equal to the permeability multiplied by the electric current enclosed in the loop.

Mathematically;

∮B.dl = I*μo

• I = current passing through the window
• μo = free space magnetic permeability

$\int B.\delta s = \mu I\\\int B.\delta s = \mu (5A - 3A)\\\int B.\delta s = 4\pi * 10^-^7 * 2A\\\int B.\delta s = 2.5*10^-^6 T.M$

The magnitude of in the path integral is 2.5*10^-6 T.M

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