Answer:
(b) 10 Wb
Explanation:
Given;
angle of inclination of magnetic field, θ = 30°
initial area of the plane, A₁ = 1 m²
initial magnetic flux through the plane, Φ₁ = 5.0 Wb
Magnetic flux is given as;
Φ = BACosθ
where;
B is the strength of magnetic field
A is the area of the plane
θ is the angle of inclination
Φ₁ = BA₁Cosθ
5 = B(1 x cos30)
B = 5/(cos30)
B = 5.7735 T
Now calculate the magnetic flux through a 2.0 m² portion of the same plane
Φ₂ = BA₂Cosθ
Φ₂ = 5.7735 x 2 x cos30
Φ₂ = 10 Wb
Therefore, the magnetic flux through a 2.0 m² portion of the same plane is is 10 Wb.
Option "b"
The magnetic flux through a 2.0 m2 portion is mathematically given as
[tex]\phi' = 10 Wb[/tex]
Question Parameters:
A uniform magnetic field makes an angle of 30o with the z-axis.
Magnetic flux through a 1.0 m2 portion of the xy plane is 5.0 Wb
Generally, the equation for the Magnetic flux is mathematically given as
[tex]\phi = BACos\theta[/tex]
Therefore
[tex]\phi = BA₁Cos\theta 5 \\\\\phi= B(1 x cos30) B \\\\\phi= 5/(cos30) B[/tex]
\phi= 5.7735 T
Therefore, the magnetic flux through a 2.0 m²
[tex]\phi' = BA₂Cos\theta\\\\\phi ' = 5.7735 x 2 x cos30[/tex]
[tex]\phi' = 10 Wb[/tex]
In conclusion, the magnetic flux through a 2.0 m²
[tex]\phi' = 10 Wb[/tex]
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