Respuesta :
Answer:
The average emf that will be induced in the wire loop during the extraction process is 37.9 V
Explanation:
The average emf induced can be calculated from the formula
[tex]Emf = -N\frac{\Delta \phi}{\Delta t}[/tex]
Where [tex]N[/tex] is the number of turns
[tex]\Delta \phi[/tex] is the change in magnetic flux
[tex]\Delta t[/tex] is the time interval
The change in magnetic flux is given by
[tex]\Delta \phi = \phi _{f} - \phi _{i}[/tex]
Where [tex]\phi _{f}[/tex] is the final magnetic flux
and [tex]\phi _{i}[/tex] is the initial magnetic flux
Magnetic flux is given by the formula
[tex]\phi = BAcos(\theta)[/tex]
Where [tex]B[/tex] is the magnetic field
[tex]A[/tex] is the area
and [tex]\theta[/tex] is the angle between the magnetic field and the area.
Initially, the magnetic field and the area are pointed in the same direction, that is, [tex]\theta = 0^{o}[/tex]
From the question,
B = 1.5 T
and radius = 15.0 cm = 0.15 m
Since it is a circular loop of wire, the area is given by
[tex]A = \pi r^{2}[/tex]
∴ [tex]A = \pi (0.15)^{2}[/tex]
[tex]A = 0.0225\pi[/tex]
∴[tex]\phi_{i} = (1.5)(0.0225\pi)cos(0^{o} )[/tex]
[tex]\phi_{i} = (1.5)(0.0225\pi)[/tex]
( NOTE: [tex]cos (0^{o}) = 1[/tex] )
[tex]\phi_{i} = 0.03375\pi[/tex] Wb
For [tex]\phi_{f}[/tex]
The field pointed upwards, that is [tex]\theta = 90^{o}[/tex]. Since [tex]cos (90^{o}) = 0[/tex]
Then
[tex]\phi_{f} = 0[/tex]
Hence,
[tex]\Delta \phi = 0- 0.03375\pi[/tex]
[tex]\Delta \phi = - 0.03375\pi[/tex]
From the question
[tex]\Delta t = 2.8 ms = 2.8 \times 10^{-3} s[/tex]
Here, [tex]N[/tex] = 1
Hence,
[tex]Emf = -N\frac{\Delta \phi}{\Delta t}[/tex] becomes
[tex]Emf = -(1)\frac{-0.03375\pi}{2.8 \times 10^{-3} }[/tex]
[tex]Emf = 37.9 V[/tex]
Hence, the average emf that will be induced in the wire loop during the extraction process is 37.9 V.
