✔ 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.