Question

A uniform magnetic field of strength 22.3 mT passes through the center of a circular coil...

A uniform magnetic field of strength 22.3 mT passes through the center of a circular coil with a radius of 5.08 cm at an angle of 30.2° relative to the area vector of the coil. What is the magnetic flux through the coil?

Homework Answers

Answer #1

magnetic flux through the coil is given by,

= B*A*cos

here,

B = magnetic field = 22.3 mT = 22.3*10^-3 T

A = area of coil = pi*r^2

r = radius of coil = 5.08 cm = 0.0508 m

= angle between area vector and magnetic field = 30.2 deg

So,

= (22.3*10^-3)*(pi*0.0508^2)*cos 30.2 deg

= 1.56*10^-4 Wb

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