Skip to content

Question

A rod of length 0.35 m moves at 2.5 m/s in a 0.2 T field perpendicular to its length. What is the
induced emf?

Options

Choose one · Correct answer highlighted

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 = 0.2 × 0.35 × 2.5 = 0.175 V . Using Φ = B A cosθ, e = -N dΦ/dt = -N A dB/dt = B l v = N B A ω sinωt, L = μ₀ N²A/l, M = e/(dI/dt)

Discussion

Comments

0 comments

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