Question

Por una espira de 0.5 m2 de área circula una corriente de 5 A. Calcula la densi- dad de flujo magnético B considerando que la espira considera la permeabilidad del medio es la del aire

Answer 1

Answer:

$\beta=B=8.05\mu T$

Explanation:

The density of the magnetic flux is given by the following formula:

$\beta=\frac{\Phi_B}{A}=\frac{ABcos\theta}{A}=Bcos\theta$

The normal vector A and the vector of the magnitude of the magnetic field are perpendicular, then, the angle is zero:

The magnitude of the magnetic field is calculated by using the formula for B at a distance of x to a point in the plane of the loop:

$B=\frac{\mu_oIR^2}{2(x^2+R^2)^{3/2}}$

For x = 0 you have:

$B=\frac{\mu_oIR^2}{2R^3}=\frac{\mu_oI}{2R}$

R is the radius of the circular loop and its values is:

$R=\sqrt{\frac{A}{\pi}}=\sqrt{\frac{0.5m^2}{\pi}}=0.39m$

Then, you replace in the equation for B with mu_o = 4\pi*10^-7 T/A:

$B=\frac{(4\pi*10^{-7}T/A)(5A)}{2(0.39m)}=8.05*10^{-6}T=8.05\mu T$

and the density of the magnetic flux is

$\beta=B=8.05\mu T$