Skip to content

Question

A rectangular loop of sides 30 cm and 12 cm moves out of a 0.6 T field at 1 m/s perpendicular to the
shorter side. What is the motional emf?

Options

Choose one · Correct answer highlighted

Explanation

**AC generator** emf e = N B A ω sin ωt, maximum when coil plane parallel to field, zero when perpendicular, time duration of emf when loop moves out of field t = L/v, L side along motion, v speed, e = B l v while cutting. For loop 0.28×0.14 m B=0.3 T v=1.4 m/s along longer side 0.28 m, cutting side 0.14 m, e=0.3×0.14×1.4=0.0588 V, duration t=0.28/1.4=0.2 s, emf exists only during exit. ε = B l v , l = 0.3 m . ε = 0.6 × 0.3 × 1 = 0.18 V . Using Φ = B A cosθ, e = -N

Discussion

Comments

0 comments

No comments yet. Be the first to start the discussion.