更新1:
a = g.sin(A), why?
an e.m question
2010-09-23 12:10 am
1. A conductor XY of mss m slides from rest down a pair of frictionless rails which are separted by a distance r and inclined at an angle A to the horizontal. A uniform magnetic field of flux density B acts vertically upwards. Deduce an equation for the e.m.f. induced in the conductor XY in terms of B, r, A, g and s, where s is the displacement of the conductor rom its rest position and g is the acceleration of gravity.
回答 (2)
2010-09-28 6:13 pm
✔ 最佳答案
One more parameter is required: The resistance of the conductor XY.Also, suppose that the rails are perfectly conducting, then with reference to the diagrams below:
圖片參考：http://i388.photobucket.com/albums/oo325/loyitak1990/Sep10/CrazyEM2.jpg
Note: It is rather to relate the induced e.m.f. with the displacement s since to find s, we have to integrate v w.r.t. time t first. Afterwards, for certain value of s, we need to evaluate for t which is rather complicated and then we need to sub it back to the expression of v.
參考: Myself
2010-09-23 4:44 am
Induced emf E of a moving conductor in a magentic field is given by,
E = B.r.v
where v is the velocity of the conductor.
Since the conductor starts from rest, use equation of motion: v^2 = u^2 + 2.a.s
with u = 0 m/s, a = g.sin(A),
hence, v^2 = 2.g.s.sin(A)
Therefore, E = B.r.square-root[2gs.sin(A)]
E = B.r.v
where v is the velocity of the conductor.
Since the conductor starts from rest, use equation of motion: v^2 = u^2 + 2.a.s
with u = 0 m/s, a = g.sin(A),
hence, v^2 = 2.g.s.sin(A)
Therefore, E = B.r.square-root[2gs.sin(A)]
收錄日期: 2021-04-29 17:38:35
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20100922000051KK00657