Answer:
The value of path integral is 2.2 [tex]\mu T.m[/tex]
Explanation:
Given:
Current carrying by long wire [tex]I = 2.5[/tex] A
Area of rectangle [tex]A = 18.75[/tex] [tex]m^{2}[/tex]
Angle with surface normal [tex]\theta =[/tex] 45°
According to the ampere's circuital law,
[tex]\int\limits {B} \, ds = \mu _{o} I_{net}[/tex]
Where [tex]\mu _{o} = 4\pi \times 10^{-7}[/tex] [tex]ds=[/tex] area element
Here, [tex]I _{net} = I \cos 45[/tex]
[tex]I_{net} = 2.5 \times \frac{1}{\sqrt{2} }[/tex]
Put value of current in above equation,
[tex]\int\limits {B} \, ds = 4\pi \times 10^{-7} \times 2.5 \times \frac{1}{\sqrt{2} }[/tex]
[tex]\int\limits {B} \, ds = 22.2 \times 10^{-7}[/tex]
[tex]\int\limits {B} \, ds = 2.2 \times 10^{-6}[/tex]
[tex]\int\limits {B} \, ds = 2.2[/tex] [tex]\mu T.m[/tex]
Therefore, the value of path integral is 2.2 [tex]\mu T.m[/tex]
