Practice question
Question
A rectangular loop of 0.24 m × 0.38 m moves out of a 0.3 T field at 0.6 m/s along its shorter side.
What is the emf?
Explanation
**Rectangular loop moving out of field** emf e = B l v, l side perpendicular to motion cutting field lines, e constant while partially in field, zero when fully out, duration t = (side parallel to motion)/v. For 0.38×0.55 m loop B=0.65 T v=0.7 m/s along shorter side 0.38 m, l=0.55 m (side cutting), e=0.65×0.55×0.7=0.25 V, lasts t=0.38/0.7=0.54 s. ε = B l v , l = 0.38 m . ε = 0.3 × 0.38 × 0.6 = 0.0684 V ≈ 0.068 V . Using Φ = B A cosθ, e = -N dΦ/dt = -N A dB/dt = B l v = N B
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