A circular loop of radius 0.10m is rotating in a uniform magnetic field of 0.20T. Find magnetic
flux B F through the loop and the magnetic field vector in Figure 1
a) Are parallel
b) Are perpendicular
c) Are at 30o
(Ans: 0, 2.0x10-3Wb, 1.0x10-3Wb )
I need the working for the questions.
To find the magnetic flux through the circular loop, we can use the formula:
Φ = B * A * cos(θ)
Where:
Φ is the magnetic flux,
B is the magnetic field strength,
A is the area of the loop, and
θ is the angle between the magnetic field vector and the plane of the loop.
Given:
B = 0.20 T (uniform magnetic field)
r = 0.10 m (radius of the circular loop)
a) When the magnetic field vector and the plane of the loop are parallel (θ = 0°), the angle θ is 0°, and cos(θ) is equal to 1.
So, for case a), the magnetic flux Φ will be:
Φ = B * A * cos(0°) = B * A = (0.20 T) * (π * r²) = 0.20π * (0.10 m)² = 0.020π Tm²
b) When the magnetic field vector and the plane of the loop are perpendicular (θ = 90°), the angle θ is 90°, and cos(θ) is equal to 0.
So, for case b), the magnetic flux Φ will be:
Φ = B * A * cos(90°) = B * A * 0 = 0
c) When the magnetic field vector and the plane of the loop form an angle of 30° (θ = 30°), we need to calculate the new area of the loop projected onto the plane perpendicular to the magnetic field vector.
The projected area A' is given by: A' = A * cos(θ).
So, for case c), the magnetic flux Φ will be:
Φ = B * A' = B * A * cos(θ)
Φ = (0.20 T) * [(π * r²) * cos(30°)]
Φ = (0.20 T) * [(π * (0.10 m)²) * cos(30°)]
Φ = (0.20 T) * [(π * 0.01 m²) * (√3/2)]
Φ = 0.020π * (0.01) * (√3/2) Tm²
So, the answers are:
a) Φ = 0.020π Tm² = 0.0628 Tm² (approximately)
b) Φ = 0 Tm²
c) Φ = 0.020π * (0.01) * (√3/2) Tm² ≈ 1.0 × 10⁻³ Tm²
To find the magnetic flux through the loop, we can use the formula:
Φ = B * A * cos(θ)
Where:
Φ is the magnetic flux,
B is the magnetic field strength,
A is the area of the loop,
and θ is the angle between the magnetic field vector and the normal to the loop.
The loop is circular, so we can find the area A using the formula:
A = π * r^2
Where:
r is the radius of the loop.
a) When the magnetic field vector and the normal to the loop are parallel:
In this case, θ = 0° Since the cosine of 0° is 1, the formula for magnetic flux becomes:
Φ = B * A * cos(0°)
= B * A
Substituting the given values:
B = 0.20 T
r = 0.10 m
A = π * (0.10 m)^2
= π * (0.01 m^2)
Φ = 0.20 T * π * (0.01 m^2)
= 0.002 T*m^2
= 2.0 x 10^(-3) Wb
b) When the magnetic field vector and the normal to the loop are perpendicular:
In this case, θ = 90° Since the cosine of 90° is 0, the formula for magnetic flux becomes:
Φ = B * A * cos(90°)
= 0
c) When the magnetic field vector and the normal to the loop are at 30°:
In this case, θ = 30° The cosine of 30° is √(3)/2, the formula for magnetic flux becomes:
Φ = B * A * cos(30°)
= B * A * √(3)/2
Substituting the given values:
B = 0.20 T
r = 0.10 m
A = π * (0.10 m)^2
= π * (0.01 m^2)
Φ = 0.20 T * π * (0.01 m^2) * √(3)/2
= 0.001 T*m^2 * √3
= 1.0 x 10^(-3) Wb
So the magnetic flux through the loop and the magnetic field vector in Figure 1 are:
a) The magnetic flux through the loop is 2.0 x 10^(-3) Wb and the magnetic field vector and the normal to the loop are parallel.
b) The magnetic flux through the loop is 0 Wb and the magnetic field vector and the normal to the loop are perpendicular.
c) The magnetic flux through the loop is 1.0 x 10^(-3) Wb and the magnetic field vector and the normal to the loop are at 30°.