Lenz's law. An induced current has a direction such that the magnetic field due to the current opposes the change in the magnetic flux that induces the current. the direction of an induced emf is that of the induced current B decreasing
Lenz’s Law: An induced current has a direction such that the magnetic field due to the current opposes the change in the magnetic flux that induces the current.The direction of an induced emf is that of the induced current. B m decreasing n dl
楞次定律: 磁通量的变化感应电流 闭合回路中产生的感应电流具有确定 的方向,总是使感应电流所产生的通 过回路面积的磁通量,去补偿或反抗 引起感应电流的磁通量的变化。 试用楞次定律判断上例中感应电动势和 感应电流的方向
楞次定律: 磁通量的变化 感应电流 闭合回路中产生的感应电流具有确定 的方向，总是使感应电流所产生的通 过回路面积的磁通量，去补偿或反抗 引起感应电流的磁通量的变化。 试用楞次定律判断上例中感应电动势和 感应电流的方向
The general means by which we can change the magnetic flux through a coil or loop: 1)Change the magnitude of b; B 2)Change the area of the coil or loop(for example, by expanding the coil or sliding it in or out of the field) 3)Change the angle between the direction of B and the area of the coil
The general means by which we can change the magnetic flux through a coil or loop: B m 1)Change the magnitude of ; B 2)Change the area of the coil or loop(for example, by expanding the coil or sliding it in or out of the field); 3)Change the angle between the direction of and the area of the coil. B
Example12-1:如图所示,棒ab长为,沿两平行的轨 道以速度v在均匀的磁场中运动,求回路中的感应电动 势 0/ B 解:(1)选回路方向 abcda;c b (2)设t时刻da=x,计算 磁通量: d ①n()= Bcos0bx… (3)应用 Faraday's law,有: dqm(t = lOcos e at d =lv cos (4)感应电动势的大小为 lvb e,方向b→>a
Example 12-1:如图所示，棒ab长为，沿两平行的轨 道以速度v在均匀的磁场中运动，求回路中的感应电动 势。 B a b c d x 解：（1）选回路方向abcda; (2)设t时刻 da=x,计算 磁通量： m (t) = xBcos （3）应用 Faraday’s law,有： = − = dt d (t ) m （4）感应电动势的大小为 vBcos ，方向 b →a 。 − dt = −vBcos dx Bcos
Example122:如图所示,棒ab长为,沿两角形的轨 道以速度v在均匀的磁场中运动,求回路中的感应电动 势 B 解:(1)选回路方向abda; BHBEEEDDE 6 (2)设t时刻da=x,计算 磁通量: d n (t)=BSMab cos 6 =-Btga cos a=- But 6 2 2 (3)应用 Faraday's1aw,有: ④n(t) =一 t -Btv2 tga cos b(4)方向:b→a
Example 12-2:如图所示，棒ab长为，沿两角形的轨 道以速度v在均匀的磁场中运动，求回路中的感应电动 势。 B a b c d x 解：（1）选回路方向abda; (2)设t时刻 da=x,计算 磁通量： m (t ) = BSabd cos （3）应用 Faraday’s law,有： = − = dt d (t ) m （4）方向：b →a = = Btg cosx Bv t tg cos Btv t g cos −