A circular loop of area 0.05 m2 rotates in a uniform magnetic field of 0.2 T. If the loop rotates about its diameter which is perpendicular to the magnetic field, find the flux linked with the loop when its plane is normal to the field.
Correct answer: A. 0.01 Wb
- A. 0.01 Wb
- B. 0 Wb
- C. 8.66 x 10^-3 Wb
- D. 0.86 Wb
Explanation
Since the plane is normal (90), the area vector will be ( 90 - 90 ) = 0 with the field. B (magnetic field) = 0.2 TA (area) = 0.05 cm2 = 0According to formula Flux will be = B A cos = 0.2 0.05 cos (0)Note cos (0) = 1 = 0.01 Wb
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About Electromagnetism
Magnetic flux density and magnetic flux describe the strength of a magnetic field and the field passing through a surface. A charged particle moving through a magnetic field experiences a force perpendicular to its velocity and may follow circular or helical motion, depending on the angle between velocity and field.
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