Respuesta :
Answer:
The magnitude of the average induced emf is 22.43 V.
Explanation:
Given that,
Magnetic field = 5.00 T
Diameter = 2.00\times10^{-2} m
Current [tex]I= 70.0\ mu s[/tex]
We need to calculate the area
Using formula of area
[tex]A=\pi r^2[/tex]
[tex]A=3.14\times(\dfrac{2.00\times10^{-2}}{2})^2[/tex]
[tex]A=0.000314\ m^2[/tex]
We need to calculate the induced emf
Using formula of induced emf
[tex]\epsilon=A\dfrac{dB}{dt}[/tex]
[tex]\epsilon=0.000314\times\dfrac{5.00-0}{70.0\times10^{-6}}[/tex]
[tex]\epsilon=22.43\ V[/tex]
Hence, The magnitude of the average induced emf is 22.43 V.
Emf is a result of changes in the magnetic flux passing through it. The magnitude of the average induced emf around this region of the brain during the treatment is 22.43 V.
What is induced emf?
It's the production of a potential difference in a coil as a result of changes in the magnetic flux passing through it.
When the flux coupling with a conductor or coil changes, electromotive Force, or EMF, is said to be induced.
The given data in the problem is;
B is the megnetic field= 5.00 T
d is the diameter= 2.00 × 10-2 m
I is the value of current= 70.0 μs.
The induced emf id is given by the formula;
[tex]\epsilon = A\frac{dB}{dt} \\\\ \epsilon = 0.000314\frac{5.00-0}{70\times 10^{-6}}\\\\ \epsilon =22.43\ V[/tex]
Hence the magnitude of the average induced emf around this region of the brain during the treatment is 22.43 V.
To learn more about the induced emf refer to the link;
https://brainly.com/question/4721624
