Question

Note: -

A finite wire lies from (-2,4,4) to (-2,4,-8) carrying a current of 6 mA. Calculate the
magnetic field intensity at P(4,4,0).​

Answer 1

Answer:

Note: -

A finite wire lies from (-2,4,4) to (-2,4,-8) carrying a current of 6 mA. Calculate the

magnetic field intensity at P(4,4,0).Note: -

A finite wire lies from (-2,4,4) to (-2,4,-8) carrying a current of 6 mA. Calculate the

magnetic field intensity at P(4,4,0).Note: -

A finite wire lies from (-2,4,4) to (-2,4,-8) carrying a current of 6 mA. Calculate the

magnetic field intensity at P(4,4,0).

Answer 2

Given:

Finite wire lies from A(-2,4,4) to B(-2,4,-8)

Current = 6mA

To find:

Magnetic field intensity at point P(4,4,0)

Solution:

A finite wire is lying between the point A(-2,4,4) to point B(-2,4,-8)

Current of the finite wire, I = 6mA = 6 × 10⁻³A

So, the magnetic field intensity at point P(4,4,0) due to finite wire is ,

 $H = \frac{I}{4\pi R} \left [ sin\theta _{1} + sin\theta _{2} \right ]$

here, $sin\theta _{1}= \frac{AP'}{AP} = \frac{4}{\sqrt{4^2 + 12^2}} = \frac{4}{\sqrt{52}}$

         $sin\theta _{2}= \frac{BP'}{BP} = \frac{8}{\sqrt{8^2 + 12^2}} = \frac{8}{10}}$

therefore, $H = \frac{6 \times 10^-^3}{4\pi 6} \left [ \frac{4}{\sqrt{52}} + \frac{8}{10} \right ]$

     H = 0.1078 mA/m

so, the magnetic field intensity is 0.1078mA/m.