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2012-05-23 20:36:43 補充：
(a) The max magnetic flux through the coil EFGH when it is completely inside the magnetic field should be 0.04^2 x 0.3 wb = 4.8 x 10^-4 wb.
(b) From t =0 s to 0.02 s
Rate of change of flux = (4.8x10^-4)/0.02 wb/s = 0.024 wb/s
From t=0.02 s to 0.1 s
Rate of change of flux = (4.8x10^-4 - 4.8x10^-4)/(0.1-0.02) wb/s = 0 wb/s
From t = 0.1 s to 0.12 s
Rate of change of flux = (0 - 4.8x10^-4)/(0.12-0.1) wb/s = -0.024 wb/s
(c) Induced emf, and hence current, only occurs during the t=0 to 0.02 s and t = 0.1 to 0.12 s time intervals. When the coil EFGH is complletely inside the magnetic field during t = 0.02 to 0.1 s, there is NO induced current, becuase the rate of change of flux is zero.
During t=0 to 0.02 s, induced emf = 0.024 v
hence, induced current = 0.024/6 A = 0.004 A (or 4 mA)
The sign is +ve (above the x-axis) as the current flows, by Lenz's Law, in the anti-clockwise direction.
During t=0.1 to 0.12 s
induced emf = 0.024 v, hence induced current = 0.024/6 A = 0.004 A
The sign is -ve (i.e. below the x-axis), as the current flows in the clockwise direction.
(d) Energy is only dissipated when the coil EFGH is moving in and moving out from the magnetic field, i.e. during the two time intervals t=0 to 0.02 s and t = 0.1 to 0.12 s, because an induced current is flowing during these two periods.
Total energy disspaited = 2 x (0.004^2 x 6 x 0.02) J = 3.84x10^-6 J
(e) The dissipated energy comes from the work done by an external force pulling the coil into and out from the magnetic field.
(optional answer)
When the side EF first enters into the filed, the anticlcokwise flowing induced current produces a magentic force in direction opposite to the motion of the coil. Hence, an external force, acting in the direction of the coil's motion, is needed to maintain the motion of the coil. This external force does work and which is dissipated in the coil by the induced current as heat.
The same argument applies when the coil is moving out from the magnetic field.
