Answer:
Φ = 0.47 Wb
Explanation:
Given that
Vrms = 115V
f= 55 Hz
The maximum voltage V given as
[tex]V_{max}= \sqrt2\ V_{rms}[/tex]
The maximum current I given as
[tex]I=\dfrac{V_{max}}{2\pi fL}[/tex]
Maximum magnetic flux
Φ = I L
[tex]\hi=\dfrac{\sqrt2\ V_{rms}}{{2\pi f}}[/tex]
Now by putting the values
[tex]\hi=\dfrac{\sqrt2\ \times 115}{{2\pi \times 55}}[/tex]
Φ = 0.47 Wb
The maximum magnetic flux through an inductor connected to a standard electrical outlet is 0.465 Wb.
Magnetic flux refers to term that describes measurement of the total magnetic field which passes through a given area.
The maximum magnetic flux can be calculated using the formula below;
Ф[tex]=[1.41 V( rms) /2\pi f][/tex].[tex].[/tex]
We were given Frequency f= 55 Hz
V( rms) = 125v
Then if we substitute the values, we have;
Ф[tex]= [ 1.41 * 125] / [2\pi *55][/tex]
Ф[tex]= 0.465 Wb.[/tex]
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