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A circular coil that has N = 290 N=290 turns and a radius of r = 11.5 cm r=11.5 cm lies in a magnetic field that has a magnitude of B 0 = 0.0705 T B0=0.0705 T directed perpendicular to the plane of the coil. What is the magnitude of the magnetic flux Φ B ΦB through the coil?

Respuesta :

Answer:

The magnitude of magnetic flux is 0.849 Wb.                

Explanation:

Given that,

Number of turns in a coil, N = 290

Radius of the coil, r = 11.5 cm    

Magnetic field, B = 0.0705 T

The magnetic field is directed perpendicular to the plane of the coil. The magnetic flux is given by :

[tex]\phi =NBA\\\\\phi =NB\pi r^2\\\\\phi =290\times 0.0705 \times \pi (11.5\times 10^{-2})^2\\\\\phi=0.849\ Wb[/tex]

So, the magnitude of magnetic flux is 0.849 Wb.                                                              

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