Question

Calculate the flux of the vector field F⃗ =−6i⃗ +5x2j⃗ −5k⃗ , through the square of side 8 in the plane y=1, centered on the y-axis, with sides parallel to the x and z axes, and oriented in the positive y-direction. Flux =__________?

Answer 1

Answer:

The flux is 682.6 Wb.

Explanation:

Given that,

Vector field $F=-6i+5x^2j-5k$

We need to calculate the flux

Using formula of flux

$\phi=\int_{-4}^{4}\int_{-4}^{4}(F\cdot j\ dxdz)$

Put the value into the formula

$\phi=\int_{-4}^{4}\int_{-4}^{4}(-6i+5x^2j-5k)1\ dxdz$

$\phi=\int_{-4}^{4}\int_{-4}^{4}(5x^2)dxdz$

$\phi=2(\dfrac{x^3}{3})_{-4}^{4}\times(z)_{-4}^{4}$

$\phi=682.6\ Wb$

Hence, The flux is 682.6 Wb.