Respuesta :
I guess the problem is asking for the induced emf in the coil.
Faraday-Neumann-Lenz states that the induced emf in a coil is given by:
[tex]\epsilon = -N \frac{\Delta \Phi}{\Delta t} [/tex]
where
N is the number of turns in the coil
[tex]\Delta \Phi[/tex] is the variation of magnetic flux through the coil
[tex]\Delta t[/tex] is the time interval
The coil is initially perpendicular to the Earth's magnetic field, so the initial flux through it is given by the product between the magnetic field strength and the area of the coil:
[tex]\Phi_i = BA=(5.30 \cdot 10^{-5}T)(10.5 \cdot 10^{-4} m^2)=5.57 \cdot 10^{-8} Wb[/tex]
At the end of the time interval, the coil is parallel to the field, so the final flux is zero:
[tex]\Phi_f = 0[/tex]
Therefore, we can calculate now the induced emf by using the first formula:
[tex]\epsilon = -N \frac{\Delta \Phi}{\Delta t}=- (250) \frac{5.57 \cdot 10^{-8} Wb - 0}{3.10 \cdot 10^{-2} s} = -4.5 \cdot 10^{-4} V[/tex]
Faraday-Neumann-Lenz states that the induced emf in a coil is given by:
[tex]\epsilon = -N \frac{\Delta \Phi}{\Delta t} [/tex]
where
N is the number of turns in the coil
[tex]\Delta \Phi[/tex] is the variation of magnetic flux through the coil
[tex]\Delta t[/tex] is the time interval
The coil is initially perpendicular to the Earth's magnetic field, so the initial flux through it is given by the product between the magnetic field strength and the area of the coil:
[tex]\Phi_i = BA=(5.30 \cdot 10^{-5}T)(10.5 \cdot 10^{-4} m^2)=5.57 \cdot 10^{-8} Wb[/tex]
At the end of the time interval, the coil is parallel to the field, so the final flux is zero:
[tex]\Phi_f = 0[/tex]
Therefore, we can calculate now the induced emf by using the first formula:
[tex]\epsilon = -N \frac{\Delta \Phi}{\Delta t}=- (250) \frac{5.57 \cdot 10^{-8} Wb - 0}{3.10 \cdot 10^{-2} s} = -4.5 \cdot 10^{-4} V[/tex]