91Ó°ÊÓ

• The current through loop 1 is$i_1=5.0A$
• The current through loop 2 is$i_2=3.0A$

Ampere’s law states that,

$\mathbf{âˆ\circledR }\stackrel{\mathbf{→}}{\mathbf{B}}\mathbf{·}\mathit{d}\stackrel{\mathbf{→}}{\mathbf{s}}\mathbf{=}{\mathit{μ}}_{\mathbf{0}}\mathit{i}$

The line integral in this equation is evaluated around a closed-loop called an Amperian loop.The current ion the right side is the net current encircled by the loop.

From the givenfig,we can findthe net current through the loop for paths1and2. Then by using Ampere’s law, we can find thevalue of integral$\mathbf{âˆ\circledR }\stackrel{\mathbf{→}}{\mathbf{B}}\mathbf{·}\mathit{d}\stackrel{\mathbf{→}}{\mathbf{s}}$for paths1and2.

According to Ampere’s law,

$âˆ\circledR \stackrel{→}{B}.d\stackrel{→}{s}={\mathrm{μ}}_{0}i$

Where the integral is around a closed loop and$i$is the net current through the loop.

For path 1, using Ampere’s law, theintegral is

$âˆ\circledR \stackrel{→}{B}.d\stackrel{→}{s}={\mathrm{μ}}_{0}i$

But from fig 29-68, the net current through the loop is

$i=-i_1+i_2$

Substituting the current in Ampere’s law, we get

$$\begin{gathered} âˆ\circledR \stackrel{→}{B}.d\stackrel{→}{s}={\mathrm{μ}}_0\left(-i_1+i_2\right) \\ =\left(4\mathrm{Ï€}×{10}^{-7}T·m/A\right)\left(-5.0A+3.0A\right) \\ =\left(4\mathrm{Ï€}×{10}^{-7}T·m/A\right)\left(-2.0A\right) \\ =-2.5×{10}^{-6}T·m \end{gathered}$$

Therefore, the value of$âˆ\circledR \stackrel{→}{B}.d\stackrel{→}{s}$ for path 1 is$-2.5×{10}^{-6}T·m$

For path 2,using Ampere’s law, theintegral is

$âˆ\circledR \stackrel{→}{B}.d\stackrel{→}{s}={\mathrm{μ}}_{0}i$

But, from fig 29-63, the net current through the loop is

$$\begin{gathered} i=-i_1-i_1-i_2 \\ =-2i_1-i_2 \end{gathered}$$

Substituting the current in Ampere’s law, we get

$$\begin{gathered} âˆ\circledR \stackrel{→}{B}.d\stackrel{→}{s}={\mathrm{μ}}_0\left(-2i_1-i_2\right) \\ =\left(4\mathrm{Ï€}×{10}^{-7}T·m/A\right)\left(-2\left(5.0A\right)-3.0A\right) \\ =\left(4\mathrm{Ï€}×{10}^{-7}T·m/A\right)\left(-13.0A\right) \\ =-1.6×{10}^{-5}T·m \end{gathered}$$

The value of the$âˆ\circledR \stackrel{→}{B}.d\stackrel{→}{s}$ for path 2 is$-1.6×{10}^{-5}T·m$