Question

Two large, parallel, current-carrying plates are oriented horizontally and the vertical distance between them is 5.0 mm. The current per unit width in each plate is 150 A/m , and both currents are in the positive x direction.

Determine the magnitude of the magnetic force per unit area exerted by the lower plate on the upper plate.

Answer 1

Answer:

    force per unit area  = 0.0141 N/m²

Explanation:

given,

vertical distance = 5.0 mm

current per unit width = 150 A/m

force

$\vec{dF} = I(d\vec{L}\times \vec{B})$

now,

$\int \vec{B} dL = \mu_0 I_{inclosed}$

$B\int_0^{2L}dL =\mu_0 K\ L$

$B\times 2L = \mu_0 K\ L$

$B = \dfrac{\mu_0}{2}\times K \hat{z}$

$\vec{dF} = I(d\vec{L}\times \dfrac{\mu_0}{2}\times K \hat{z}})$

$\vec{dF} = K x (\dfrac{\mu_0}{2}\times K \times d\vec{L}\times \hat{z})$

$\vec{dF} =\dfrac{\mu_0}{2}\times K^2 (x \times z)$

$\dfrac{\vec{dF}}{ (x \times z)} =\dfrac{\mu_0}{2}\times K^2$

                                              =  $\dfrac{4\pi \times 10^{-7}}{2}\times (150)^2$

            force per unit area  = 0.0141 N/m²