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Difference between revisions of "Motor Effect"
 
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|-
 
|-
 
|[[File:FlemingsLeftHandRule.png|center|300px]]
 
|[[File:FlemingsLeftHandRule.png|center|300px]]
 +
|[[File:MotorEffect.png|center|300px]]
 
|-
 
|-
−
| style="height:20px; width:200px; text-align:center;" |
+
| style="height:20px; width:200px; text-align:left;" |
 
<math>\overrightarrow{B}</math>: First finger [[Magnetic Field|field]].
 
<math>\overrightarrow{B}</math>: First finger [[Magnetic Field|field]].
  
−
<math>\overrightarrow{I}</math>:  Second finger [[Electrical Current|current]].
+
<math>{I}</math>:  Second finger [[Electrical Current|current]].
  
 
<math>\overrightarrow{F}</math>: The [[force]].
 
<math>\overrightarrow{F}</math>: The [[force]].
 +
| style="height:20px; width:200px; text-align:left;" |The direction on the [[force]] on this [[wire]] can be found using [[Fleming's Left Hand Rule]].
 
|}
 
|}
  
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<math>F = BIl</math>
 
<math>F = BIl</math>
  
−
Where:
+
Where
  
 
<math>B</math> = The [[Magnetic Flux Density]] (strength of [[Magnetic Field|magnetic field]]).
 
<math>B</math> = The [[Magnetic Flux Density]] (strength of [[Magnetic Field|magnetic field]]).
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===Example Calculations===
 
===Example Calculations===
 +
{| class="wikitable"
 +
| style="height:20px; width:300px; text-align:center;" |A [[wire]] carrying a [[Electrical Current|current]] of 2.3A is inside a 55mT [[Magnetic Field|magnetic field]] that extends along the [[wire]] by a [[length]] of 6.1cm. Calculate the [[force]] on the [[wire]] correct to two [[Significant Figures|significant figures]].
 +
| style="height:20px; width:300px; text-align:center;" |23cm of [[wire]] carrying 1.2A of [[Electrical Current|current]] is contained in a 37mT [[Magnetic Field|magnetic field]]. Calculate the [[force]] on the [[wire]] correct to two [[Significant Figures|significant figures]].
 +
|-
 +
| style="height:20px; width:300px; text-align:left;" |'''1. State the known quantities in [[SI Unit]]s.'''
 +
 +
<math>B</math> = 55mT = 55x10<sup>-3</sup>T
 +
 +
<math>I</math> = 2.3A
 +
 +
<math>l</math> = 6.1cm = 6.1x10<sup>-2<sup>m
 +
 +
| style="height:20px; width:300px; text-align:left;" |'''1. State the known quantities in [[SI Unit]]s.'''
 +
 +
<math>B</math> = 37mT = 37x10<sup>-3</sup>T
 +
 +
<math>I</math> = 1.2A
 +
 +
<math>l</math> = 23cm = 23x10<sup>-2<sup>m
 +
|-
 +
| style="height:20px; width:300px; text-align:left;" |'''2. [[Substitute (Maths)|Substitute]] the numbers into the [[equation]] and [[Solve (Maths)|solve]].'''
 +
 +
<math>F = BIl</math>
 +
 +
<math>F = 55\times 10^{-3} \times 2.3 \times 6.1 \times 10^{-2}</math>
 +
 +
<math>F = 7.7165 \times 10^{-3}N</math>
 +
 +
<math>F \approx 7.7\times 10^{-3}N</math>
 +
 +
| style="height:20px; width:300px; text-align:left;" |'''2. [[Substitute (Maths)|Substitute]] the numbers into the [[equation]] and [[Solve (Maths)|solve]].'''
 +
 +
<math>F = BIl</math>
 +
 +
<math>F = 37\times 10^{-3} \times 1.2 \times 23 \times 10^{-2}</math>
 +
 +
<math>F = 1.0212 \times 10^{-2}N</math>
 +
 +
<math>F \approx 1.0\times 10^{-2}N</math>
 +
|}
 +
 +
===References===
 +
====AQA====
 +
 +
:[https://www.amazon.co.uk/gp/product/1471851370/ref=as_li_tl?ie=UTF8&camp=1634&creative=6738&creativeASIN=1471851370&linkCode=as2&tag=nrjc-21&linkId=01c69b0ae058f809cf636033e6ba793e ''Motor effect, page 228, GCSE Physics, Hodder, AQA '']
 +
:[https://www.amazon.co.uk/gp/product/0008158770/ref=as_li_tl?ie=UTF8&camp=1634&creative=6738&creativeASIN=0008158770&linkCode=as2&tag=nrjc-21&linkId=ec31595e720e1529e49876c3866fff6e ''Motor effect, page 249, GCSE Physics; Student Book, Collins, AQA '']
 +
:[https://www.amazon.co.uk/gp/product/019835939X/ref=as_li_tl?ie=UTF8&camp=1634&creative=6738&creativeASIN=019835939X&linkCode=as2&tag=nrjc-21&linkId=57e96876985fc39b1a3d8a3e3dc238b6 ''Motor effect, pages 220-221, GCSE Physics; Third Edition, Oxford University Press, AQA '']
 +
:[https://www.amazon.co.uk/gp/product/1782946403/ref=as_li_tl?ie=UTF8&camp=1634&creative=6738&creativeASIN=1782946403&linkCode=as2&tag=nrjc-21&linkId=32a0abb60dff015b15b50e9b1d7b4644 ''Motor effect, pages 221-227, GCSE Combined Science Trilogy; Physics, CGP, AQA '']
 +
:[https://www.amazon.co.uk/gp/product/1471851362/ref=as_li_tl?ie=UTF8&camp=1634&creative=6738&creativeASIN=1471851362&linkCode=as2&tag=nrjc-21&linkId=7d78d70a2044ee9982dae010c94af92a ''Motor effect, pages 281, GCSE Combined Science Trilogy 2, Hodder, AQA '']
 +
:[https://www.amazon.co.uk/gp/product/1782945970/ref=as_li_tl?ie=UTF8&camp=1634&creative=6738&creativeASIN=1782945970&linkCode=as2&tag=nrjc-21&linkId=a120d24dcc7cc7a58192069a3aafc1d2 ''Motor effect, pages 295-302, GCSE Physics; The Complete 9-1 Course for AQA, CGP, AQA '']
 +
:[https://www.amazon.co.uk/gp/product/178294558X/ref=as_li_tl?ie=UTF8&camp=1634&creative=6738&creativeASIN=178294558X&linkCode=as2&tag=nrjc-21&linkId=f0dfb66dafcb0c6e9449e7b1a4ae1ac320 ''Motor effect, pages 94, 95, GCSE Physics; The Revision Guide, CGP, AQA '']
 +
 +
====Edexcel====
 +
 +
:[https://www.amazon.co.uk/gp/product/1292120223/ref=as_li_tl?ie=UTF8&camp=1634&creative=6738&creativeASIN=1292120223&linkCode=as2&tag=nrjc-21&linkId=068ecf40278c32406a7f1c6e66751417 ''Motor effect, page 172, GCSE Physics, Pearson Edexcel '']
 +
:[https://www.amazon.co.uk/gp/product/1782945741/ref=as_li_tl?ie=UTF8&camp=1634&creative=6738&creativeASIN=1782945741&linkCode=as2&tag=nrjc-21&linkId=30da4f2178da182547b62a7329d13b57 ''Motor effect, page 197, GCSE Combined Science; The Revision Guide, CGP, Edexcel '']
 +
:[https://www.amazon.co.uk/gp/product/1292120193/ref=as_li_tl?ie=UTF8&camp=1634&creative=6738&creativeASIN=1292120193&linkCode=as2&tag=nrjc-21&linkId=572df39392fb4200db8391d98ae6314e ''Motor effect, page 406, GCSE Combined Science, Pearson Edexcel '']
 +
:[https://www.amazon.co.uk/gp/product/1782948163/ref=as_li_tl?ie=UTF8&camp=1634&creative=6738&creativeASIN=1782948163&linkCode=as2&tag=nrjc-21&linkId=0fdbfd5dd397d6e24a9dfb250f08587f ''Motor effect, pages 273-278, GCSE Physics, CGP, Edexcel '']
 +
:[https://www.amazon.co.uk/gp/product/1782945733/ref=as_li_tl?ie=UTF8&camp=1634&creative=6738&creativeASIN=1782945733&linkCode=as2&tag=nrjc-21&linkId=2a2dbec9db6bf5766c0458d908fa0a52 ''Motor effect, pages 87, 88, GCSE Physics; The Revision Guide, CGP, Edexcel '']

Latest revision as of 18:20, 22 January 2023

Contents

Key Stage 4

Meaning

The motor effect is the force on a current carrying wire in a magnetic field.

About The Motor Effect

When an a wire has an electrical current it has a magnetic field. If this wire is in the presence of an external magnetic field the two fields will interact causing a force.
The magnitude of the force depends upon:
The force on a current carrying wire is at right angles to both the current and the magnetic field.
Fleming's Left Hand Rule can be used to find the direction of the force.
 
 

\(\overrightarrow{B}\): First finger field.

\({I}\): Second finger current.

\(\overrightarrow{F}\): The force.

The direction on the force on this wire can be found using Fleming's Left Hand Rule.

Equation

Force = (Magnetic Flux Density) x (Current) x (Length)

\(F = BIl\)

Where

\(B\) = The Magnetic Flux Density (strength of magnetic field).

\(I\) = The Electrical Current through the wire.

\(l\) = The length of wire inside the magnetic field.

\(F\) = The force on the wire.

Example Calculations

A wire carrying a current of 2.3A is inside a 55mT magnetic field that extends along the wire by a length of 6.1cm. Calculate the force on the wire correct to two significant figures. 23cm of wire carrying 1.2A of current is contained in a 37mT magnetic field. Calculate the force on the wire correct to two significant figures.
1. State the known quantities in SI Units.

\(B\) = 55mT = 55x10-3T

\(I\) = 2.3A

\(l\) = 6.1cm = 6.1x10-2m

1. State the known quantities in SI Units.

\(B\) = 37mT = 37x10-3T

\(I\) = 1.2A

\(l\) = 23cm = 23x10-2m

2. Substitute the numbers into the equation and solve.

\(F = BIl\)

\(F = 55\times 10^{-3} \times 2.3 \times 6.1 \times 10^{-2}\)

\(F = 7.7165 \times 10^{-3}N\)

\(F \approx 7.7\times 10^{-3}N\)

2. Substitute the numbers into the equation and solve.

\(F = BIl\)

\(F = 37\times 10^{-3} \times 1.2 \times 23 \times 10^{-2}\)

\(F = 1.0212 \times 10^{-2}N\)

\(F \approx 1.0\times 10^{-2}N\)

References

AQA

Motor effect, page 228, GCSE Physics, Hodder, AQA
Motor effect, page 249, GCSE Physics; Student Book, Collins, AQA
Motor effect, pages 220-221, GCSE Physics; Third Edition, Oxford University Press, AQA
Motor effect, pages 221-227, GCSE Combined Science Trilogy; Physics, CGP, AQA
Motor effect, pages 281, GCSE Combined Science Trilogy 2, Hodder, AQA
Motor effect, pages 295-302, GCSE Physics; The Complete 9-1 Course for AQA, CGP, AQA
Motor effect, pages 94, 95, GCSE Physics; The Revision Guide, CGP, AQA

Edexcel

Motor effect, page 172, GCSE Physics, Pearson Edexcel
Motor effect, page 197, GCSE Combined Science; The Revision Guide, CGP, Edexcel
Motor effect, page 406, GCSE Combined Science, Pearson Edexcel
Motor effect, pages 273-278, GCSE Physics, CGP, Edexcel
Motor effect, pages 87, 88, GCSE Physics; The Revision Guide, CGP, Edexcel
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