VCE PHYSICS AOS 1 UNIT 4. Presented by Nick Howes

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1 VCE PHYSICS AOS 1 UNIT 4 Presented by Nick Howes

2 Forces on wires and DC Motors Key knowledge calculate magnitudes, including determining the directions of, and magnetic forces on, current carrying wires, using F = ni B where the directions of I and B are either perpendicular or parallel to each other investigate and explain the operation of simple DC motors consisting of: one coil, containing a number of loops of wire, which is free to rotate about an axis two magnets providing a uniform magnetic field a commutator a DC power supply 2

Forces in a wire A current carrying wire in magnetic field can experience a force The size of the force depends on the orientation of the field, size of the field

3 Forces in a wire A current carrying wire in magnetic field can experience a force The size of the force depends on the orientation of the field, size of the field and size of the current The source of this force can be attributed to the fact that a current produces a magnetic field which can interact with the external field 3

4 Size of force When current parallel to field zero force is experienced. Considering perpendicular field to current flow only: The size of the force F = n B I L N = number of wires B = strength of field (Tesla) I = current (Amperes) L = length of wire in the field (metres) Note proportionality of each quantity to size of force [Also, F = n B I L sin θ - not examinable numerically] 4

5 Concept Explorer What is the magnitude of the force exerted on a bunch of 15 wires, each carrying a current of 1.5 Amperes and are of length 0.7 m, running perpendicular to a magnetic field of strength 350 mt? A N B. 5.5 N C N D. 368 N 5

6 Worked solution What is the magnitude of the force exerted on a bunch of 15 wires, each carrying a current of 1.5 Amperes and are of length 0.7 m, running perpendicular to a magnetic field of strength 350 mt? Worked Response Known values B = 350 x 10-3 T, L = 0.7 m, n = 15 turns, I = 1.5 A. Use F = nbil = 15 x 350 x 10-3 x 1.5 x 0.7 = 5.51 N 6

7 Direction of Force #3 Right Hand Slap Rule Using a flat (right) hand, point fingers in direction of the field and thumb in the direction of the current in the wire. Palm points in direction of the force experienced. 7

8 Concept Explorer What is the direction of the force acting on the wire marked I in the diagram below A. Out the page B. Into the page C. Up the page D. Down the page 8

9 Electron beams Electron beams obey this rule but need to remember that electron flow is opposite to conventional current flow 9

10 DC Motor Essentially 2 wires in a uniform field experiencing equal and opposite forces. Each side obeys RHS rule and as current flow is opposite so is direction of force Results in torque (not strictly in AOS) Consider one side at a time when analysing 10

11 Concept Explorer In the diagram below, which of the 2 options below should would result in the motor rotating in the direction shown? A. Connect x to positive and y to negative B. Connect y to positive and x to negative 11

Split ring commutator To maintain rotation either the field or the current must be reversed when the motor s field become parallel to the applied field split ring

12 Split ring commutator To maintain rotation either the field or the current must be reversed when the motor s field become parallel to the applied field split ring commutator used to reverse current Purpose: Reverses the current every half turn to maintain direction of torque Care when describing direction of force on each side 12

13 Ways to increase turning effect Using an understanding of F = nbil and Torque the following can be deduced: Greater current Longer wire in field (more turns) Stronger field Wider motor 13

14 Quiz The magnet shown produces a uniform magnetic field of 3.0 x 10-2 T between the pole pieces. The coil JKLM, wound on a square armature of side 0.50 m consists of 20 turns of wire. There is a current of 2.0 A in each turn of the coil. The armature can rotate about the axis XY. For the coil oriented horizontally, as, calculate the magnitude of the total force exerted on the 20 turns of side JK. A. 0.6 N B N C N D. 2.4 N 14

Quiz With the same set up as before, identify the best response below for the size of the force in a single wire between K and L. The magnet shown produces a uniform magnetic field of 3.

15 Quiz With the same set up as before, identify the best response below for the size of the force in a single wire between K and L. The magnet shown produces a uniform magnetic field of 3.0 x 10-2 T between the pole pieces. The coil JKLM, wound on a square armature of side 0.50 m consists of 20 turns of wire. There is a current of 2.0 A in each turn of the coil. The armature can rotate about the axis XY. For the coil oriented horizontally, as, calculate the magnitude of the total force exerted on the 20 turns of side KL. A. 0.6 N B N C N D. 0.0 N 15

16 Quiz Figure below shows a power line that carries a DC current of 2000 A running from west to east. The earth s magnetic field at this site is T, running horizontally from south to north. An engineer is concerned about the electromagnetic force due to the earth s magnetic field on the wire between the two support poles, which are 20 m apart. Calculate the magnitude and direction (north, south, east, west, up, down) of the force due to the earth s magnetic field on the 20 m section of wire. A. 16 N up B. 16 N Down C. 1.6 N Up D. 1.6 N Down 16

17 The copyright in substantial portions of this material is owned by the Victorian Curriculum and Assessment Authority. Used with permission. The VCAA does not endorse this product and makes no warranties regarding the correctness or accuracy of its content. To the extent permitted by law, the VCAA excludes all liability for any loss or damage suffered or incurred as a result of accessing, using or relying on the content. Current and past VCAA exams and related content can be accessed directly at 17

Note that a current-carrying solenoid produces a dipole field similar to that of a bar magnet. The field is uniform within the coil.

Note that a current-carrying solenoid produces a dipole field similar to that of a bar magnet. The field is uniform within the coil. An electrical current produces a magnetic field that is directed around it. Conventional current is the flow of positive charge. Hence, it is directed from the positive terminal of the power supply, through