MAGNETIC FIELDS
Question 1
SMF 1 to MFP c
SMF 2 to MFP a
SMF 3 to MFP b
Diagram 1 is of a coil, we use coils to become 'bar' magnets, so the field around a coil must look like the field around a bar magnet. Mfp 3 is the best option.
The next easiest one to deal with is '3'. Don't feel constrained to do the questions in the order that they are written, sometimes it is easier to do them in another order.
So '3' is a straight current carrying conductor, so the filed is given by the direction that the fingers curl if the thumb is in the direction of the current. This means that they are concentric circles, the direction of the field is controlled by the direction of the current (which is not shown)
That means that '2' must behave like two versions of '3'. The field lines will be in the opposite direction if the direction of the current was given, then you could show this. The non-concentric circles are a result of the interaction between the two fields.
FORCES ON WIRES
Question 1
Force
= BIl
= 0.5 3.0 2.0
= 3.0 N
Question 2 ANS A
The current is flowing from left to right. The field is into the page. The Right Hand Slap Rule shows that the force is up the page.
Question 3
A
The right hand grip rule specifies the direction of the field. This is covered better in dot point 3.
DC MOTOR
Question 1
F = nBIl = 50 x 0.005 x 3.0 x 0.05 = 0.038 N (remember to convert cm to m)
Question 2
The direction of the force is always perpendicular to both P and the magnetic field, hence response B
Question 3
The commutator needs to be free to rotate since this is the way that the connection is made between the external current source and the coil itself via the brushes rubbing on the commutator.
He motor rotates since the force on the sides of the coil that are perpendicular to the magnetic applies a torque to the coil which is down on the side nearest the south pole and up on the side nearest the north pole, given the direction of current flow shown on the diagram. However, when the coil rotates into the vertical position, the force on side P at the bottom needs to reverse in order for the torque to be applied in the clockwise direction. For this to happen, the direction of the current needs to reverse and it is the role of the split-ring commutator to do this. The split in the ring is such that as the coil gets to the vertical position, the direction of the current flow through the coil reverses and the direction of the torque applied to the coil remains in the same direction.
Question 4 B
The magnetic field points from left to right (north to south). Therefore, the current must flow in at Z and out at X (right hand grip rule)
Question 5
Magnitude of the force
= BIl = 0.10 1.5 0.12 = 0.018 N
Question 6 A
Using the right hand slap rule (fingers – field, palm – push, thumb – current), the armature will start to rotate anticlockwise
(side JK moves down)
Question 7
H
The right hand slap rule give that the fingers are pointing form N to S (ie. From right to left), the current is pointing to the top of the page, so the palm is facing upwards. Therefore the force is out of the page.
Question 8
Which choice (A-I in the answer key) best indicates the direction of the magnetic force on the wire at point y?
Question 8
I
The force at the point y is zero, because the direction of the current is parallel to the field here. Therefore the component of the current perpendicular to the field is zero.
Question 9
Which choice (A-I in the answer key) best indicates the direction of the magnetic force on the wire at point z?
Question 9
G
The right hand slap rule give that the fingers are pointing form N to S (ie. From right to left), the current is pointing to the bottom of the page (in general), so the palm is facing downwards. Therefore the force is into the page.
Question 10
Which choice (A-I in the answer key) best indicates the direction of the magnetic force on the wire at point w?
Question 10
H
The right hand slap rule give that the fingers are pointing form N to S (ie. From right to left), the current is pointing to the top of the page, so the palm is facing upwards. Therefore the force is out of the page.
In order to convert the arrangement in Figure 5 into a motor, Thomas provides an axis for rotation. He realises that there must be current flowing through the coil when it is rotating, so he attaches a set of slip rings that rotate with the coil as illustrated in Figure 5.
The coil is initially set with its plane parallel to the magnetic field as shown, and the switch is open so that no current flows.
Question 11
The switch is closed. Which one of the statements (A-D) best describes the situation after current begins to How?
A. The coil begins to rotate, but stops after turning through 90°.
B. The coil begins to rotate and, after rotating one half turn, rotates back to its original position. It then continues to oscillate in this way.
C. The coil does not move. D. The coil rotates continuously.
Question 11
A (or B)
The best answer is A, because when the switch is closed the current will flow around the loop. The magnetic field will exert a force on this current and cause the loop to rotate. When the loop has rotated through 900 , the direction of the force is now non acting to rotate the loop, but to dilate it. If we assume that there is some frictional forces, then the loop will stop. If we assume that there are no frictional forces then the loop will oscillate. Eventually the loop would come to rest, because there will be some friction acting.
Question 12
If the slip rings used by Thomas in the circuit of Figure 5 were replaced with a commutator, which one of the statements (A-D) best describes the situation after the switch is closed, and the current begins to flow?
A. The coil begins to rotate, but stops after turning through 90°.
B. The coil begins to rotate and, after rotating one half turn, rotates back to its original position. It then continues to oscillate in this way.
C. The coil does not move. D. The coil rotates continuously.
Question 12
D
The best answer is D, because the function of the commutator is to reverse the direction of the current every half cycle. This will mean that the direction of the force changes every half cycle. This will cause the coil to rotate continuously.
Two physics students set up a 50-turn coil (JKLM), which is free to rotate about the axis shown as the dashed line below. The loop is placed between the poles of a magnet, in a uniform magnetic field of 0.040 T. The current in the coil is 1.5 A.
Question 13
What is the magnitude of the magnetic force on side JK (length = 0.050 m) of the 50-turn coil, when oriented as shown above?
Question 13
Note that there are now 50 turns in the coil, so the force will be given by F = nBiL.
F = nBil
F = 50 × 0.040 × 1.5 × 0.050
F = 0.15 N
Question 14
What is the magnitude of the magnetic force on side KL (length = 0.040 m) of the 50-turn coil, when oriented as shown above?
Question 14
Zero
Because the wire KL is parallel to the field, so there isn't a force acting on it. A force only acts when the wire is perpendicular (or has a component that is perpendicular) to the field. As side KL is oriented parallel to the magnetic field it experiences zero force.
The students now turn off the current and set the coil at rest, oriented as shown below. They then turn the current on again.
Question 15
What happens to the coil after the current is turned on? Explain your answer.
Question 15
If the coil is at rest in this position when the current is turned on, the coil will not rotate.
Consider JK. The field (your fingers) is from left to right. The current (your thumb) is pointing into the page. This means that your palm (the force) is downwards.
If you consider ML, then the field is still from left to right, but the current is out of the page. This means that the force on ML is upwards.
Consider KL. Field to the right, current down the page, so the force is out of page.
Consider JM. Field to the right, current up the page, so the force is into the page.
The effect is to compress all the sides, because there are forces acting 'inwards' on all sides. There is no torque, so the coil will not spin. In a simple motor like this, it will not start from this position, but if it wasn't 'oriented' exactly in this position then there would be a torque acting and the coil would begin the spin. It would usually have enough momentum to get past this point and it the current changed it would continue to spin.
A single loop of wire in a uniform magnetic field is shown below. The loop can rotate, and is shown at
three different orientations. In each case there is a current flowing around the coil from W to X to Y to Z.
Question 16
The magnetic field is 0.10 T, and the current in the loop is 0.30 A. With the loop in orientation (a), what is the magnitude of the force acting on side WX of the coil? The length of side WX is 0.030 m. Show your working.
Question 16
F = nBIL
F = 1 × 0.010 × 0.30 × 0.30
F = 0.0009 N
= 9 × 10-4 N
Below, the arrows indicate possible directions of the force on side WX of the loop in the three orientations (a), (b) and (c). The arrows in each orientation are in a plane perpendicular to the axis of rotation of the loop.
Question 17
For each orientation below, circle the head of the arrow which best represents the direction of the magnetic force on side WX of the coil. If there is no force on the side, write NF under the diagram.
Question 17
a) pointing to the leftb) pointing to the leftc) pointing to the left
(a)The field is down, the current is into the page, so the thumb points into the page with the fingers pointing down. This makes the palm of your hand face left, so the force is in the direction pointing to the left
(b)The field is down, the current is into the page, so the thumb points into the page with the fingers pointing down. This makes the palm of your hand face left, so the force is in the direction pointing to the left
(c)The field is down, the current is into the page, so the thumb points into the page with the fingers pointing down. This makes the palm of your hand face left, so the force is in the direction pointing to the left
There isn't any change in either 3 orientations, the field was constant, and the wire had a current going into the page in these three situations, so each answer is identical. Don't get fussed that on the exam, they actually gave you such a simple question. You need to solve every problem on its merits.
In Figure 6 below, the arrows indicate possible directions of the force on side XY for the loop in orientations (a) and (c) shown in Figure 4.
Question 18
For each of the two orientations in Figure 6, circle the head of the arrow which best represents the direction of the magnetic force on side XY of the coil. If there is no force on the side, write NF under the diagram.
Question 18
a) NFb) Into the plane of the page
(a)So the section of wire XY is not at right angles to the field. It is parallel to the field, so this means that the force acting on it will be zero.
(b)Now the wire is at right angles to the field, so there will be a force acting on it. The field is down the page (direction of your fingers) the current is from left to right across the page (direction of your thumb). So your palm must be facing into the page (you need a flexible wrist for this). So the force is into the plane of the page.
Figure 5 below shows four positions (A, B, C, D) of the coil of a DC motor. The coil can be assumed to be a single wire which is in a uniform magnetic field parallel to the coil when in the orientation shown in diagram A.
It rotates in the direction indicated, about the axis which passes through the middle of sides LM and NK. The coil is attached to a commutator, to which current is passed by brushes (not shown in the figure).
Figure 5
Question 19
For the coil as shown in orientation A of Figure 5, in which direction is the current flowing in the side KL?
Explain your answer.
Question 19
Place your fingers in the direction of the field, from left to right. For the loop to rotate clockwise the force on KL must be upwards. So have your palm facing up. This means that your thumb points from K - L.
So the current flows from K to L.
Side KL of the coil is 0.10 m long, and a magnetic force of 0.60 N acts on it.
Question 20
If the magnetic field has a magnitude of 1.5 T, what is the magnitude of the current in the coil?
Question 20
F = nBIL
0.60 = 1 × 1.5 × I × 0.1
I = 4 amp
Question 21
Consider two cases:
a. The coil is at rest with the orientation shown in diagram A of Figure 5.
b. The coil is at rest with the orientation shown in diagram B of Figure 5.
In the answer book explain what would happen, in each case, if current is allowed to flow in the coil. Your answer should discuss the forces on each side of the coil, and their net effect.
Question 21
- If the coil starts in this position, the forces acting on KL and MN will spin the loop in the direction shown, because of the torques acting See answer 234.
- If the coil starts in this position the forces acting on KL and MN will tend to try to squeeze or expand the loop, but not spin it.
EMF
A 2.0 m long wire is moved down the page out of the magnetic field (of strength 0.5T ) at a velocity of 2.0 ms-1 as shown in the diagram below.
Question 1
Calculate the emf generated in the wire.
Question 1
EMF = vBl = 2.0 0.5 2.0 = 2.0 V
Question 2
Describe two different ways in which the emf generated could be doubled.
Question 2
Either double the velocity or double the magnetic flux intensity or double the length of the wire. (any two of these)
Question 3
Do induced currents in a circuit oppose or magnify the change in magnetic flux that induces them? Fully explain your answer in detail.
Question 3
This is Lenz’s Law. The induced current flows in a direction such that the magnetic flux produced opposes the change in magnetic flux that produced the current. This is conservation of energy. If the reverse were true, energy would be created out of nothing.
A square loop of wire of side 0.01 m is moved through a magnetic field.
The maximum magnetic flux generated is 5.0 x 10-4 Wb.
Question 4
Calculate the magnetic flux density of the magnetic field.
Question 4
Maximum magnetic flux = BA
The square loop is moved at a constant speed through the magnetic field in one direction and then back again through the magnetic field in the reverse direction.
The graph showing the magnetic flux through the coil as a function of time is shown below.
Question 5
Draw a graph which shows the variation of emf through the loop against time. Explain the main features of your graph.
Question 5
The induced emf is the gradient of the magnetic flux-time graph. The gradient changes from constant positive to constant negative and then again to constant positive to constant negative.
The graph is shown below. (5 marks)
Jackie and Jim are studying electromagnetic induction. They have a small permanent magnet and a coil of wire wound around a hollow cylinder as shown below.
Jackie moves the magnet through the coil in the direction shown at constant speed.
Question 6
Indicate on the diagram the direction of the induced current that flows in the resistor. Explain the physics reason for your choice.
Question 6
Use Lenz’s Law and RH grip rule
From Lenz’s Law the induced field produced by the coil must oppose the increasing external magnetic field caused by the magnet moving closer. The induced field must therefore flow to the right towards the magnet.
From the RH grip rule, to get a magnetic field that points out of the centre of the coil to the right will require the current to flow anti-clockwise when viewed from the magnet towards the end of the coil.
The current through the external resistor will thus flow from left to right
They next decide to move the magnet, at a constant speed, all the way through the coil and out the other side.
Question 7
Which one of the diagrams (A–D) below best shows how the current through the coil varies with time?
Question 7
As the magnet approaches the induced current will increase to a maximum as the magnet enters the coil, it will then be constant while the magnet moves through the coil (it will not be zero as the magnet is still moving). As the magnet exits the coil and moves away this will result in a reduction of the field within the coil and the induced current will flow in the direction needed to try and maintain the field ie it will reverse.
Diagram D is the only one that fits this reasoning
The on-off action of a lightning stroke produces an electromagnetic field surrounding the stroke. It is this field that causes the crackling in your radio or TV during a thunderstorm.
The magnetic field,B, produced by a lightning stroke varies with time as shown in Figure 1.