KBC 發問於 科學及數學其他 - 科學 · 8 年前

# physic 難題 29

Please let me know whether it's wrong for my attached solutions part (a) & (c)

I don't know how to calculate the part (b) and part (d), does it calculate wrong??

In part (e), where has the dissipated energy? is it for heat energy??

### 1 個解答

• 天同
Lv 7
8 年前
最愛解答

Your question cannot be seen using your given links. User ID and password are required to open the JPEG images.

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.