Induced emf

cipker

2010-07-24 6:52 am

A uniform magnetic field of induction B fills a cylindrical volume of radius R. A metal rod of length L is placed as shown. If B is changing at a rate of dB/dt, show that the emf that is produced by the changing magnetic field and that acts between the ends of the rod is given by:
E = dB/dt (L/2)[R^2-(L/2)^2]^(1/2)

圖片參考：http://imgcld.yimg.com/8/n/HA00140412/o/701007230178513873371730.jpg

更新1:

Do you mean: If we consider the loop of the triangle, the magnetic field lines will not cut the opposite sides of the triangle if the field is increasing or decreasing. It is because the field lines will only move along the radii (the opposite sides of the triangle) when the field is changing.

更新2:

Then, the induced emf in the loop will only be that induced in the rod. In this way, we can apply faraday's law to the area of the loop.

回答 (1)

天同

2010-07-24 8:43 am

✔ 最佳答案

Let O be the centre of the circle where the magnetic field is acting. Then by joining O to the two ends of the rod, an isosceles triangle is formed. This triangle thus has a base of length L and the two sides each of length R.

Hence, height of triangle = square-root[R^2-(L/2)^2]

Magnetic flux within this triangle = B.Awhere A is the area of the triangle.Hence A = (1/2).L.{square-root[R^2-(L/2)^2]}
By Faraday's Law, induced emf E = d(B.A)/dt
i.e. E = A/(dB/dt) = (dB/dt).(L/2).square-root[R^2-)L/2)]^2
The physical process is that when the magnetic flux density is changing (either increasing or decreasing), it can be construed as the field lines are moving radially inward (for increasing field) towards or outward (for decreasing field) away from the centre of the circle, so as to decrease or increase respectively the spacing between field lines. As such, only field lines bound by the above triangular area would be cut by the metal rod, causing an induced emf.

2010-07-24 11:50:51 補充：
Yes...you are right. We could assume during a field change, field lines only move along the radial direction (i.e. along the radii). In this way, only field lines bound by the triangle is of significance.

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