Respuesta :
(a) The magnetic flux through the coil is given by
[tex]\Phi = N B A \cos \theta[/tex]
where
N is the number of turns in the coil
B is the intensity of the Earth's magnetic field
A is the area enclosed by one coil
[tex]\theta[/tex] is the angle between the direction of B and the perpendicular to A.
Initially, the coil is perpendicular to the Earth's magnetic field, therefore [tex]\theta=0^{\circ}[/tex] and [tex]\cos \theta=1[/tex], so the magnetic flux through the coil is given by:
[tex]\Phi_i = NBA = (200) (5.20 \cdot 10^{-5} T)(10.8 \cdot 10^{-4} m^2)=1.12 \cdot 10^{-5}Wb[/tex]
(b) After the coil is rotated, its plane is parallel to the magnetic field. This means that [tex]\theta=90^{\circ}[/tex] and [tex]\cos \theta=0[/tex], so the magnetic flux through the coil is simply zero:
[tex]\Phi_f = NBA (0) = 0[/tex]
(c) The variation of magnetic flux through the coil is
[tex]\Delta \Phi = \Phi_f - \Phi_i = 0-(1.12 \cdot 10^{-5} Wb) = -1.12 \cdot 10^{-5} Wb[/tex]
we can ignore the negative sign since we are interested only in the magnitude of the average emf, not in its direction.
For Faraday-Neumann-Lenz law, the average emf induced in the coil is given by
[tex]\epsilon = \frac{\Delta \Phi}{\Delta t}= \frac{1.12 \cdot 10^{-5} Wb}{5.00 \cdot 10^{-2}s}=2.25 \cdot 10^{-4} V [/tex]
[tex]\Phi = N B A \cos \theta[/tex]
where
N is the number of turns in the coil
B is the intensity of the Earth's magnetic field
A is the area enclosed by one coil
[tex]\theta[/tex] is the angle between the direction of B and the perpendicular to A.
Initially, the coil is perpendicular to the Earth's magnetic field, therefore [tex]\theta=0^{\circ}[/tex] and [tex]\cos \theta=1[/tex], so the magnetic flux through the coil is given by:
[tex]\Phi_i = NBA = (200) (5.20 \cdot 10^{-5} T)(10.8 \cdot 10^{-4} m^2)=1.12 \cdot 10^{-5}Wb[/tex]
(b) After the coil is rotated, its plane is parallel to the magnetic field. This means that [tex]\theta=90^{\circ}[/tex] and [tex]\cos \theta=0[/tex], so the magnetic flux through the coil is simply zero:
[tex]\Phi_f = NBA (0) = 0[/tex]
(c) The variation of magnetic flux through the coil is
[tex]\Delta \Phi = \Phi_f - \Phi_i = 0-(1.12 \cdot 10^{-5} Wb) = -1.12 \cdot 10^{-5} Wb[/tex]
we can ignore the negative sign since we are interested only in the magnitude of the average emf, not in its direction.
For Faraday-Neumann-Lenz law, the average emf induced in the coil is given by
[tex]\epsilon = \frac{\Delta \Phi}{\Delta t}= \frac{1.12 \cdot 10^{-5} Wb}{5.00 \cdot 10^{-2}s}=2.25 \cdot 10^{-4} V [/tex]
A) The magnitude of the magnetic flux (φ_initial) through the coil before it is rotated is; φ_i = 1.123 × 10⁻⁵ Wb
B) The magnitude of the magnetic flux (φ_final) through the coil after it is rotated is; φ_F = 0 Wb
C) The magnitude of the average emf induced in the coil is;
ε = 22.5 × 10⁻⁵ Wb
We are given;
Number of turns; N = 200 turns
Area; A = 10.8 cm² = 0.00108 m²
Magnitude of magnetic field; B = 5.2 × 10⁻⁵ T
Time; t = 5 × 10⁻² s = 0.05 s
A) Formula for the magnitude of the magnetic flux (φ_initial) through the coil before it is rotated is;
φ_i = NAB
Plugging in the relevant values gives;
φ_i = 200 × 0.00108 × 5.2 × 10⁻⁵
φ_i = 1.123 × 10⁻⁵ Wb
B) Formula for the magnitude of the magnetic flux (φ_final) through the coil after it is rotated is;
φ_f = NAB cos θ
After it is rotated, θ = 90°
Thus;
φ_f = 200 × 0.00108 × 5.2 × 10⁻⁵ × cos 90°
φ_f = 0 Wb
C) The magnitude of the average emf induced in the coil is;
ε = -(φ_f - φ_i)/t
Thus;
ε = -(0 - 1.123 × 10⁻⁵)/0.05
ε = 22.46 × 10⁻⁵ Wb
To 3 significant figures;
ε = 22.5 × 10⁻⁵ Wb
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