Electromagnetic Induction Question

1. The Figure below shows a wire coil being squeezed in a uniform magnetic field.

                                    

(a) When the coil is being squeezed,
      (i) explain why there is an induced e.m.f in the coil,
      (ii) state and explain the direction of induced current, clockwise or anticlockwise, in the coil.
(b) Once the coil has reached its final squeezed shape, state and explain the direction of induced current in the coil.

(a)(i) Analysis:
Recall that whenever there is an induced emf, there must be a change in magnetic field lines cutting the coil.

Answer:
When the coil is being squeezed, there is a change in the magnetic field lines cutting the coil. By Faraday's Law, an emf is induced in the coil.
(ii)Analysis:
Think about rectangular coil in the a.c generator. To find out the direction of induced current, we have to analyze it part by part.

By Fleming’s Right hand Rule, the direction of induced current (middle finger) is perpendicular to both the direction of magnetic field from North to South (index finger) and the motion (thumb). Therefore, the direction of induced current on the left is upwards while that on the right is downwards, and thus the induced current flows clockwise.

OR

By using right hand grip rule, the thumb represents the direction of magnetic field while other curling fingers represent the direction of induced current. Therefore, the induced current flows clockwise.
(b) Once the coil has reached its final squeezed shape, there is no change in the magnetic field lines cutting the coil, and hence no induced emf, thus, no induced current.

3. The diagram below shows a simple a.c. generator.

                                    

(a) Explain
      (i) the function of the slip-rings A and B,
      (ii) why an emf is induced in the coil as it rotates.
(b) State and explain the direction of induced current created in coil JKLM at the instant shown above.
(c) The coil of the generator rotates 100 times per second. At this speed, the maximum induced emf is 0.2 V. Sketch a graph of emf against time for a time interval of 20 ms from the instant shown above. Label your axes clearly.
(d) The ends P and Q are connected to a transformer that is 100% efficient. The diagram below shows the setup for the transformer connected to identical lamps rated at 5V, 10W each.



(i) State the type of transformer.
(ii) Explain why material A cannot be made from steel.
(iii) If both lamps are operating at optimum brightness, calculate the currents in both primary and secondary circuits, Ip and Is.

(a) Analysis:
These two questions are recall questions. You should be able to do it after browsing through the content knowledge in this website.

(i) To transfer the alternating induced current in the coil to the external circuit.
(ii) When the coil rotates, there is a change in the magnetic field lines cutting the coil, hence emf is induced inside the coil.
(b) By Fleming's Right Hand Rule, the direction of induced current (middle finger) is perpendicular to both the direction of magnetic field from North to South (index finger) and the direction of movement (thumb). Thus, the direction of induced current is anticlockwise (MLKJ).
(c) Analysis:
frequency = 100Hz, so period = 1/f = 0.01s = 10ms
Thus, for 20ms, there would be 2 cycles
Also, at this instance, the coil is in horizontal position, thus the graph starts from maximum induced emf. This means the graph is a cosine graph.

                                    

(d)(i) Step up transformer

Comment: a common mistake is to think that primary coil is always on the left, which is not necessarily true! Primary coil is associated with power source! In this case, it is on the right.

(ii) Steel is not easily magnetized and retains its magnetism permanently.
(iii) At optimum brightness, this means that pd across both lamps is 5V.
Current across each lamp = P/V = 10/5 = 2.0 A
So Is = 2 x 2 = 4.0 A

Since the transformer is 100% efficient,
      VsIs = VpIp
5.0 * 4.0 = 0.2 * Ip
        Ip = 100A

4. The ends of a long wire are connected to a bulb to make a large loop as shown in Figure below. The wire is slowly wound round a soft-iron C-core on which there is a 500-turn coil connected to the mains. The mains is labelled 240 V, 300 W. After winding about five turns of wire round the core, the bulb begins to glow; after another five turns it glows more brightly.

                

(a) Explain why the bulb lights up.
(b) Explain why more turns make the bulb brighter.
(c) Explain what would happen if an iron bridge was placed across the top of the C-core.

(a) As the coil is joined to the a.c. mains, an alternating current flows through it which produces a changing magnetic field lines. This changing magnetic field lines is linked to the secondary coil on the left. By Faraday's Law, an emf is induced and hence induced current flows and the bulb glows.
(b) The changing magnetic field lines cuts across more turns of wire. Hence there is greater induced emf and thus greater induced current flowing through the bulb.
(c) Lamp glows brighter.
More magnetic field lines are now concentrated within the soft-iron core and hence greater number of magnetic field lines cutting the wire at one time.

5. Figure 1 shows the top view of a copper rod, XY, sliding down another two copper rods which are connected to a centred-zero galvanometer. The whole setup is placed in a region with magnetic field directed as shown.
Figure 2 shows the side view of the setup.



The copper rod, XY, is released and it slide down along the other two copper rods.
(a) Indicate on the Figure 1 the direction of the induced current flow in rod XY.
(b) Indicate on the Figure 1 the induced magnetic force acting on the rod, XY. Label it as F.
(c) Compare the motion of the rod XY, with and without the presence of magnetic field. Explain your answer.
(d) If the two copper rods are tilted at an angle bigger than θ, state one change that will be observed. Explain your observation.

(a) Analysis:
The rod is moving to produce current and there is no power source. We have to use Fleming's Right Hand Rule to determine the direction of induced current. Answer is shown in part(b)
(b) Analysis:
Here, magnetic force is produced due to the induced current. Thus, we have to use Fleming's Left Hand Rule to determine its direction.

             

(c) Without magnetic field, XY will slide faster due to the gravitational force acting on it.
With magnetic field, XY will generate a magnetic force acting in the opposite direction. So it will slide slower.
(d) The galvanometer deflection will be greater.
As XY will slide faster, the rate of change of magnetic flux in it will be greater and according to Faraday’s Law, the induced emf and hence the induced current will be bigger.

6. The figure below shows part of a cathode ray tube. The cathode ray will hit the centre O of the fluorescent screen when plates A and B are switched off.

       

(a) Explain why the fluorescent screen has to be grounded.
(b) If plate A is positive in potential relative to plate B, where will the beam of cathode ray hit the screen? (explain with reference to the axes shown)
(c) If plate A is replaced by the South pole of a magnet, while B is replaced by the North pole of a magnet, where will the beam of cathode ray hit the screen? (explain with reference to the axes shown)
(d) C and D are another pair of metal plates similar to A and B but placed horizontally. The figure below shows how plates C and D are positioned when viewed front on.

             

If the beam of cathode ray hit the screen at point P, state how each of the four plates must be charged.
(e) Now, the output voltage from a transformer is connected to the Y-plates (A and B) after the necessary adjustment the trace shown below is obtained.

             

Sketch the appearance of the trace in each of the following cases.
(i) the voltage gain on the Y-plates is increases from 1.0 V/div to 2.0 V/div
(ii) the connections to the Y-plates are disconnected.
(iii) the time-base is switched off and the Y-plates are re-connected.

(a) To allow electrons to flow to earth, so that it will not obstruct incoming electrons.
(b) Analysis:
As plate A and B are charged, the electron will move towards the positively charged plate due to unlike charges attract.

Answer:
Along OY / Above point O along line Y-Y¹
(c) Analysis:
Now, plate A and B creates magnetic field at that region. When the electron moves towards the region, it will experience a magnetic force and its direction is determined by Fleming's Left Hand Rule.

Answer:
Along OX¹ / To the right of O along line X-X¹.
(d) Analysis:
As P is at third quadrant, it means the electron is attracted by both plate B and C.

Answer:
Plates B and C must be positively charged
Plates A and D must be negatively charged
(e) Analysis:
(i) When the Y-gain increases by two times, the size of trace decreases by two times or it is halved. Since the question didnot mention time base, the frequency remains the same.
(ii) No input to Y-plates means no vertical deflection of trace. However, time base is still on, the electron will still sweep horizontally across the screen.
(iii) Now, time base is off, thus, the electron only deflects up and down. A vertical line is observed.

Answer:

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