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面对题海,只有找出解题思路,总结出解题方法,才能跳出题海.在电磁感应现象中,常常遇到计算流过导体电量的问题,笔者根据自己多年的教学经验,总结出求电量的两条思路如下,以供同行参考.思路一当闭合电路中的磁通量发生变化时,根据法拉第电磁感应定律,平均感应电动势 E=N(ΔΦ)/(Δt),平均感应电流 I=E/R=N(ΔΦ)/(RΔt),则通过导体横截面的电量 q=I△t=N(ΔΦ)/R,可见电量 q 仅由线圈匝数、回路总电阻和磁通量的变化决定,与磁通量变化时间、线圈运动速率无关.例1 如图1所示,导线全部为裸导线,半径为 r的圆内有垂直圆平面的匀强磁场,磁感应强度为 B,
In the face of the sea of problems, only by finding out the problem-solving ideas and summarizing the method of solving problems can we jump out of the sea of questions. In the phenomenon of electromagnetic induction, we often encounter the problem of calculating the amount of electricity flowing through the conductor. The author summarizes it based on his own years of teaching experience. The two ideas of seeking power are as follows for reference by the peers. The idea is that when the magnetic flux in the closed circuit changes, according to Faraday’s law of electromagnetic induction, the average induced electromotive force E=N(ΔΦ)/(Δt), the average induced current I= E/R=N(ΔΦ)/(RΔt), then through the conductor cross section q=IΔt=N(ΔΦ)/R, it can be seen that the quantity q is determined only by the number of turns of the coil, the total resistance of the loop and the change of the magnetic flux. , Regardless of the flux change time, the coil movement rate has nothing to do. Example 1 As shown in Figure 1, the wire is all bare wire, a circle with a radius of r in the circle has a uniform circular magnetic field, the magnetic induction intensity is B,