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An emf of 17.0 mV is induced in a 483-turn coil when the current is changing at the rate of 10.0 A/s. What is the magnetic flux through each turn of the coil at an instant when the current is 4.30 A?

Respuesta :

fahadisahadam

Answer:

[tex]0.0000151Tm^{2}[/tex]

Explanation:

[tex]E=17.0mV=0.017V\\t=10s\\I=4.30A[/tex]

[tex]N=483 turns[/tex]

The relation to find the magnetic flux is:

φ[tex]=\frac{LI}{N}[/tex]

so we need to find L

[tex]L=\frac{E}{dI/dt}[/tex]

[tex]L=\frac{0.017}{10.0} \\L=0.0017H[/tex]

Now we can apply:

φ=[tex]\frac{L*I}{N}[/tex]

φ=[tex]\frac{0.0017*4.30}{483}[/tex]

φ=[tex]0.0000151Tm^{2}[/tex]

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