Question

In an ac generator, if a coil of \( N \) turns and area \( A \) is rotated at \( \nu \) revolutions per second in a uniform magnetic field \( B \), then the motional emf produced is equal to \[ (t = 0 s, the coil is perpendicular to the field) \]

• A. \( NBA (2\pi \nu) \sin(2\pi \nu t) \)
• B. \( NBA^2 (2\pi \nu) \sin(2\pi \nu t) \)
• C. \( N^2 B^2 A^2 (2\pi \nu) \sin(2\pi \nu t) \)
• D. \( NBA (4\pi \nu) \sin(2\pi \nu t) \)

Hint:

In ac generators, the emf produced is sinusoidal and depends on the number of turns, area of the coil, magnetic field strength, and the frequency of rotation.

Answer 1

The Correct Option is A

The emf induced in a rotating coil in a magnetic field is given by the formula: \[ \mathcal{E} = NBA (2\pi \nu) \sin(2\pi \nu t) \] Where: - \( N \) is the number of turns of the coil - \( B \) is the magnetic field strength - \( A \) is the area of the coil - \( \nu \) is the frequency of rotation - \( t \) is the time At \( t = 0 \), the coil is perpendicular to the field, so the maximum induced emf occurs at this time. The term \( (2\pi \nu) \) accounts for the angular frequency of the rotation, and the sine function represents the variation of the induced emf over time. Thus, the motional emf produced is \( \boxed{NBA (2\pi \nu) \sin(2\pi \nu t)} \).