Question

A long straight wire carrying a 2.5 A current passes through the centre of a rectangle of dimensions 2.5 m by 7.5 m at the angle of 45 degrees to the surface normal. What is the path integral of ∮B⇀ ∙ ds⇀ around the outside of the rectangle? Express your answer using two significant figures. The answer should be in μT∙m.

Answer 1

Answer:

The value of path integral is 2.2 $\mu T.m$

Explanation:

Given:

Current carrying by long wire $I = 2.5$ A

Area of rectangle $A = 18.75$ $m^{2}$

Angle with surface normal $\theta =$ 45°

According to the ampere's circuital law,

     $\int\limits {B} \, ds = \mu _{o} I_{net}$

Where $\mu _{o} = 4\pi \times 10^{-7}$ $ds=$ area element

Here, $I _{net} = I \cos 45$

 $I_{net} = 2.5 \times \frac{1}{\sqrt{2} }$

Put value of current in above equation,

  $\int\limits {B} \, ds = 4\pi \times 10^{-7} \times 2.5 \times \frac{1}{\sqrt{2} }$

  $\int\limits {B} \, ds = 22.2 \times 10^{-7}$

  $\int\limits {B} \, ds = 2.2 \times 10^{-6}$

  $\int\limits {B} \, ds = 2.2$ $\mu T.m$

Therefore, the value of path integral is 2.2 $\mu T.m$