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 × [tex]10^{-4}[/tex] 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 = [tex]\int \vec{B}.d\vec{A}[/tex] .................1
and Ф = BA cos∅ ............2
so B = [tex]\frac{\phi }{Acos\theta }[/tex]
and we know
A = ab
so
B = [tex]\frac{\phi }{abcos\theta }[/tex] ..............3
put here value
B = [tex]\frac{3.10\times 10^{-4} }{2.80 \times 10^{-2}\times 3.20 \times 10^{-2}\times cos60}[/tex]
solve we get
B = 0.692 T
We have that from the Question"" it can be said that the magnitude of the magnetic field be to produce a flux of 3.10×10−4Wb through the surface is
From the Question we are told
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?
Generally the equation for area is mathematically given as
[tex]A=2.80*10^{-2}*3.20*10^{-2}\\\\A=8.96*10^{-4}[/tex]
Therefore
Magnetic Flux is mathematically given as
[tex]BAcos\theta=3.1*10^{-4}\\\\B=\frac{3.1}{8.96*cos60}[/tex]
B=0.691T
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