Respuesta :
Answer:
[tex]\varepsilon_{prom}=51.59V[/tex]
Explanation:
1) Notation and data given
N= 80 represent the turns
B=1.5 T represent the magnetic field
Dimensions =25cm x40cm
[tex]\phi=41[/tex]° represent the angle respect to the plpane of the coil
[tex]\phi_i=90-41=49[/tex]° since we need the angle respect to the magnetic field
[tex]\phi_f =0[/tex]° since the final position is perpendicular to the field.
[tex]\Delta t= 0.08s[/tex]
[tex]\Phi_{B,f}[/tex] represent the final flux through the coil
[tex]\Phi_{B,i}[/tex] represent the initial flux through the coil
[tex]\varepsilon[/tex] represent the induced emf, known as "electromagnetic induction" and is defined as "the production of voltage in a coil because of the change in a magnetic flux through a coil" (Variable of interest).
2) Formulas to use
We can begin calculating the area given by:
[tex]A=0.25mx0.40m=0.1m^2[/tex]
We can use the formula for the average magnitude when we have an induced emf, given by:
[tex]\varepsilon_{prom}=N|\frac{N\Phi_B}{\delta t}|=N|\frac{\Phi_{B,f}-\Phi_{B,i}}{\delta t}|[/tex] (1)
We have another formula for the flux through the coil given by:
[tex]\Phi_B =BAcos(\phi)[/tex]
Replacing this into equation (1) we got:
[tex]\varepsilon_{prom}=\frac{NBA|cos(\phi_f)-cos(\phi_i)|}{\Delta t}[/tex] (2)
3) Calculate the final answer
Now we can replace all the values given into equation (2) like this:
[tex]\varepsilon_{prom}=\frac{(80)(1.5T)(0.1m^2)|cos(0)-cos(49)|}{0.0800s}=51.59V[/tex]
