Answer to Question #182695 in Electricity and Magnetism for Andrei

Question

  A current-carrying circular plate with A= 32.2 m2 is inclined at 34° with the Earth’s field direction. If the magnitude of the magnetic field is 48 125 nT, solve for the (a) angle between the normal to the surface and the B vector, and the (b) magnetic flux through the antenna

 

Answer 1

We have given,

$B = 48.125nT= 48.125\times 10^{-9}T$

$A = 32.2 m^2$

$\theta = 34^\circ$

a.) Angle between the normal to the surface and the B vector $= 90- \theta = 90-34 = 56^ \circ$

b.) We know flux $\phi$ can be given as, $\phi = BAcos\theta$

$\phi = 48.125 \times 10^{-9} \times 32.2 \times cos56^ \circ$

$\phi = 866.53 \times 10^{-9}$ Wb