Question

A horizontal rectangular surface has dimensions 2.80 cm by 3.20 cm and is in a uniform magnetic field that is directed at an angle of 30.0∘ above the horizontal. What must the magnitude of the magnetic field be to produce a flux of 3.10×10−4Wb through the surface?

Answer 1

Answer:

magnitude of the magnetic field 0.692 T

Explanation:

given data

rectangular dimensions = 2.80 cm by 3.20 cm

angle of 30.0°

produce a flux Ф = 3.10 × $10^{-4}$  Wb

solution

we take here rectangular side a and b as a = 2.80 cm and b = 3.20 cm

and here angle between magnitude field and area will be ∅ = 90 - 30

∅ = 60°

and flux  is express as

flux Ф = $\int \vec{B}.d\vec{A}$   .................1

and Ф = BA cos∅    ............2

so B = $\frac{\phi }{Acos\theta }$    

and we know

A = ab

so

B = $\frac{\phi }{abcos\theta }$    ..............3

put here value

B =  $\frac{3.10\times 10^{-4} }{2.80 \times 10^{-2}\times 3.20 \times 10^{-2}\times cos60}$  

solve we get

B = 0.692 T