Question

A horizontal rectangular surface has dimensions 2.80 cm by 4.15 cm and is in a uniform magnetic field that is directed at an angle of 32.0 ∘ above the horizontal.

What must the magnitude of the magnetic field be to produce a flux of 3.80 ××10−4−4 WbWb through the surface?

Answer 1

Magnetic flux on a rectangular loop is given by,

= B*A*cos

here, A = area of loop = l*b = 2.80*4.15 cm^2 = 2.80*4.15*10^-4 m^2

= magnetic flux = 3.80*10^-4 Wb

= angle between normal vector and magnetic field = 90.0 - 32.0 = 58.0 deg  

B = magnetic field = ??

So, B = /(A*cos)

B = (3.80*10^-4)/[(2.80*4.15*10^-4)*cos(58.0 deg)]

B = 0.617 T