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A square loop of side 0.25 m with 10 turns moves at 8 m/s perpendicular to a 0.6 T magnetic field. If the loop's resistance is 4 Ω, the induced current is:

Correct answer: B. 1.5 A

  • A. 0.75 A
  • B. 1.5 A
  • C. 2.25 A
  • D. 3 A

Explanation

Motional emf = ε = NBvl, where N = 10, B = 0.6 T, v = 8 m/s, l = 0.25 m. Thus, ε = 10 × 0.6 × 8 × 0.25 = 6 V. Current I = ε/R = 6/4 = 1.5 A.

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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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