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:left;" | | | style="height:20px; width:200px; text-align:left;" | | ||
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<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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===Example Calculations=== | ===Example Calculations=== | ||
+ | {| class="wikitable" | ||
+ | | style="height:20px; width:200px; 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:200px; 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:200px; 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:200px; 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:200px; 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:200px; 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> | ||
+ | |} |
Revision as of 10:16, 5 March 2019
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 Current - The greater the current the greater the force.
- The Magnetic Field - The greater the strength of magnetic field the greater the force.
- 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. \(\overrightarrow{I}\): 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.
\(F = BIl\) \(F = 37\times 10^{-3} \times 1.2 \times 23 \times 10^{-2}\) \(F = 1.0212 \times 10^{-2}N\) \(F \approx 1.0\times 10^{-2}N\) |