vce physics - edrolo physics aos 1 unit 4 presented by nick howes...

VCE PHYSICS AOS 1 UNIT 4

Presented by Nick Howes

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VCE Physics Unit 4 © Nick Howes & Edrolo 2015 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



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



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]



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.  0.37 N

B.  5.5 N

C.  5500 N

D.  368 N



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



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.



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



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



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



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



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



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



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.  0.03 N

C.  0.12 N

D.  2.4 N



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.  0.03 N

C.  0.15 N

D.  0.0 N



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 4.0 × 10-5 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



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vce physics - edrolo physics aos 1 unit 4 presented by nick howes...

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