phy E&M

2014-12-21 6:07 am
https://i.na.cx/N209O5.jpg
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1. Why are (2), (3) correct?


https://i.na.cx/E8N07i.jpg
https://i.na.cx/5iw9Ce.jpg
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2. part c: is the answer wrong? Direction of magnetic force should be horizontally right, not up the plane?

https://i.na.cx/mZm85s.jpg
3. part a: should it be clockwise or anticlockwise?

https://i.na.cx/5doNLx.jpg
4. why are 1,3 correct but not 2,4?

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5. why are 5,6 incorrect?

https://i.na.cx/G6cyvq.jpg
https://i.na.cx/7r6684.jpg
6. How do I kw whether it is E/V which changes when the plates are moved apart?




thanks

回答 (1)

2014-12-22 12:07 am
✔ 最佳答案
1. The current carrying coil behaves like an electromagnet, with the axis along the normal CN. The North Pole is on the top and South pole at the bottom. Since a magnet tends to align with the external magnetic field lines, the coil will move with its axis aligning with the field lines.

2.Yes, direction of magnetic force is pointing horizontally towards the right.
I suppose you should have found the induced current = 2.25 A
The magnetic force = 1.5 x 0.15 x 2.25 N = 0.5063 N
Thus the resolved component of magetic force pointing upwards along the plane = 0.5063cos(30) N = 0.4384 N, which is what given in the answer.

3. The coil should be moving clockwise.
The left brush is at a higher potential then the right one. Find the coil moving direction by Fleming's Left Hand Rule.



2014-12-21 16:07:55 補充:
sorry...word limit have been reached.

2014-12-21 16:11:16 補充:
4. Circuit 2 is wrong.
The right hand switch doesn't not connect to any "contact point". When the left hand switch is turned to the "left contact point"(the black dot), there is no complete circuit. Tthe bulb will not glow.

Circuit 3 is wrong.

2014-12-21 16:12:34 補充:
When the left hand switch is turned to the "left contact point", short-circuit will be formed.

Circuit 1 and 4 are correct.

5. Options 5 and 6 are correct.
Ouput power = Input power x transformer efficency

2014-12-21 16:14:07 補充:
Since input power is fixed, a higher efficiency leads to a higher output power (i.e. the brighter the lamp).

Option 5: reduction of eddy current leads to less heating in the iron core, thus increasing the efficiency.

2014-12-21 16:14:56 補充:
Option 6: thicker wires leads to less heating of the coils. More power is transferred to the bulb (Law of Conservation of Energy).

6. There is an ambiguity to this question. It doesn't specify if the plates are isolated or connected to a battery (fixed voltage).

2014-12-21 16:15:55 補充:
Should the plated be isolated, the charges on the plates remain the same. An increase of separation leads to a decrease of capacitance and hence an increase of voltage, without any change of electric field.

2014-12-21 16:16:23 補充:
Should the plates be kept at a fixed voltage, an increase of separation leads to a decrease of electric field.

Based on the 4 given options, it is believed that the plates are kept at fixed voltage.

2014-12-21 16:17:14 補充:
Next time, do not put so many questions in a single post.

2014-12-24 16:23:34 補充:
Your suppl questions:
1. When the coil is in equilibrium, the normal (which behaves like the axis of a magnet) should align with the magnetic field lines, which are running vertically. The plane of the coil is perpendicular to its normal, thus becomes horizontal.

2014-12-24 16:29:46 補充:
The physical reason behind is straight forward. The magnetc force acted on the coil, when its plane is horizontal, is horizontally outward from its axis. There is no vertical force (or torque) to turn the coil.

2014-12-24 16:33:04 補充:
3. Yes, you could use Fleming's Right Hand Rule instead.

2014-12-24 16:41:25 補充:
4. Sorry, circuit 1 can't work. Only circuit 4 (option 4) is the correct answer.

2014-12-24 23:18:17 補充:
Your further questions:
3. Current flows from D to C and then from F to E. The field lines are from right to left. The motion, by Fleming's Right Hand Rule, is clockwise.

2014-12-24 23:21:37 補充:
4. See my new answer to Q4 above.


收錄日期: 2021-04-15 17:38:19
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