Answer to Question #184172 in Electricity and Magnetism for Blank 0822

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

a)

$\theta=90°-\alpha=56°,$

b)

$\Phi=AB\cos \theta=AB\sin \alpha,$

$\Phi=32.2\cdot 48125\cdot 10^{-9}\cdot 0.559=0.867~mWb.$