A magnetic field is perpendicular to the plane of a single-turn circular coil. The magnitude of the field is changing, so that an emf of 0.65 V and a current of 3.8 A are induced in the coil. The wire is the re-formed into a single-turn square coil, which is used in the same magnetic field (again perpendicular to the plane of the coil and with a magnitude changing at the same rate). What emf and current are induced in the square coil?

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davidlcastiblanco davidlcastiblanco · Answer 1 · 2019-11-18T23:45:06+01:00

Answer:

[tex]E_{square}=0.51v[/tex]

[tex]i_{square}=2.98A[/tex]

Explanation:

[tex]E_{mf}=-N*\frac{d\alpha}{dt}[/tex]

[tex]E_{mf}=-N*A*\frac{d\beta}{dt}[/tex]

[tex]N=1[/tex]

[tex]E_{circle}=A_{circle}*\frac{d\beta}{dt}[/tex]

[tex]E_{square}=E_{circle}*\frac{A_{square}}{A_{circle}}[/tex]

[tex]E_{square}=E_{circle}(\frac{\frac{\pi^2}{4}*r^2}{\pi*r^2})[/tex]

[tex]E_{square}=E_{circle}*\frac{\pi}{4}[/tex]

[tex]E_{square}=0.65v*\frac{\pi}{4}[/tex]

[tex]E_{square}=0.51v[/tex]

[tex]i_{square}=i_{circle}*\frac{E_{square}}{E_{circle}}[/tex]

[tex]i_{square}=3.8A*\frac{0.51v}{0.65v}[/tex]

[tex]i_{square}=2.98A[/tex]