Answer:
Emax = 13.209V
Explanation:
Given the following data;
Magnitude of the magnetic field, B = 0.35T
Number of turns, n = 200
Frequency, f = 60Hz
Area of the coil = 5cm² (to square meter = 5 *10^-4 = 0.0005m²)
[tex] Angular frequency, w = 2 \pi f[/tex]
Substituting into the above equation, we have;
[tex] Angular frequency, w = 2 *3.142 *60[/tex]
Angular frequency, w = 377.04
We know that as the coil rotates between 0 and 1; sin(wt) would vary sinusoidally. Therefore, the maximum EMF occurs when the value of sin(wt) is equal to 1.
Mathematically, the maximum EMF is given by;
[tex] Emax = nBAw[/tex]
Where;
Substituting into the equation, we have;
[tex] Emax = 200 * 0.35 * 0.0005 * 377.4[/tex]
Emax = 13.209V
Therefore, the maximum EMF in the coil is 13.209 volts.
The maximum emf in the coil will be 13.209V.
Emf is 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 magnitude of the magnetic field,= 0.35T
n is the number of turns, = 200
f is the Frequency = 60Hz
A is the area of the coil = 5cm² = 0.0005m²
The maximum emf is given as;
[tex]\rm E_{max}= nBA\omega \\\\ \rm E_{max}= 200 \times 0.35 \times 0.005 \times 2 \pi \times 60\\\\ \rm E_{max}= 13.209 \ V[/tex]
Hence the maximum emf in the coil will be 13.209V.
To learn more about the induced emf refer to the link;
https://brainly.com/question/16764848
