A square coil of 16 cm has 200 turns and rotates in a uniform magnetic field of magnitude 0.05 T. If the peak emf is 12 V, what is the angular velocity of the coil?
Correct answer: A. 47 rad s-1
- A. 47 rad s-1
- B. 49 rad s-1
- C. 43 rad s-1
- D. 45 rad s-1
Explanation
To determine the angular velocity (ω) of the coil, we use the formula for the peak emf (E) induced in a rotating coil in a magnetic field: E = NBAω, where N is the number of turns, B is the magnetic field strength, A is the area of the coil, and ω is the angular velocity. For a square coil with a side of 16 cm, the area A = (0.16 m)² = 0.0256 m². Substituting the given values into the rearranged formula ω = E / (NBA), we have:ω = 12 V / (200 × 0.05 T × 0.0256 m²) ≈ 47 rad/s.Therefore, the correct angular velocity of the coil is approximately 47 rad/s. Options B, C, and D are incorrect because their calculations do not align with the correct application of the formula.
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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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