Physics Book - User contributions [en]

Magnetic Field of a Long Thick Wire Using Ampere's Law

2016-11-28T03:32:28Z

Jeong-Eun Moon: /* External links */

'''Claimed by Jeong-Eun Moon (Fall 2016)'''

==The Main Idea==

This section explains how to find the magnetic field near a long thick wire using Ampere's Law. Finding the magnetic field using Ampere's Law is very simple compared to finding it using the Biot-Savart law.

===Formula for Ampere's Law===

Here is the proof:

[[File:Ampere's_Law_Proof.png]]

[[File:Ampere's_Law.png]]

==How to Find Magnetic Field of A Long Thick Wire==

[[File:Wire.png]]

To find the magnetic field <math>B</math> at a distance <math>r</math> from the center of the long wire apply Ampere's Law. By the symmetry of the wire <math>B</math> will always be constant and tangential to the circular path at every point around the wire.

The magnetic field is everywhere parallel to the path for a circular path centered on wire. The direction of magnetic field can be determined by the right hand rule. The magnetic field's direction is perpendicular to the wire and is in the direction the fingers curl if you wrap the wire around. The direction of current is where your thumb points to.

[[File:Ampere's_Picture.png]]

The path integral <math>{{\oint}d\vec{l}}</math> in this situation is equal to the circumference of the circular path around the wire. This is equal to <math>2πr</math>.

Using the formula above and plugging in <math>{d\vec{l}}</math> we have: <math>{B(2πr) = μ_0I}</math>. To solve for <math>B</math> divide both sides by <math>2πr</math>.

This results in the equation: <math>{B = \frac{μ_0I}{2πr}}</math> which is equal to <math>{\frac{μ_02I}{4πr}}</math>. This is the equation for the magnetic field of a long thick wire that is found using the Biot-Savart law.

==Examples==

===Simple===

What is the magnetic field at a point 0.03 m away from a wire that has a current of 7 amperes?

Solution: <math>I=7</math> and <math>r=0.03</math>. So inserting this into the formula gives <math>{B = \frac{μ_0(7)}{2π(0.03)}}</math>. This results in 4.67e-5 T

===Middling===

A long straight wire suspended in the air carries a conventional current of 7.4 amperes in the -x direction as shown (the wire runs along the x-axis). At a particular instant an electron at location < 0, -0.004, 0 > m has velocity < -3.5 e5, -4.2 e5, 0 > m/s. What is the magnetic field due to the wire at the location of the electron?

Solution: <math>{B = \frac{μ_0(7.4)}{2π(0.004)}}</math> and use right hand rule. The answer is < 0, 0, 3.7e-4 > T.

===Hard===

Using the same values as in the middling problem calculate the magnetic force on the electron due to the wire.

Solution: The formula for magnetic force is <math>qv×B</math>. So by using the value for B computed above and calculating the cross product between B and qv we find that this equals < 2.49e-17, -2.07e-17, 0 > N. Don't forget that because this is the force on an electron q is negative.

== See also ==

See the page on Ampere's Law for a more in depth look at the law itself: [[Ampere's Law]]

For more applications of Ampere's Law see: [[Magnetic Field of a Toroid Using Ampere's Law]] and [[Magnetic Field of Coaxial Cable Using Ampere's Law]]

===Further reading===

Chabay, Sherwood. (2015). Matter and Interactions (4th ed., Vol. 2). Raleigh, North Carolina: Wiley.

===External links===

http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/magcur.html

https://www.khanacademy.org/science/physics/magnetic-forces-and-magnetic-fields/magnetic-field-current-carrying-wire/v/magnetism-6-magnetic-field-due-to-current

https://en.wikibooks.org/wiki/Electrodynamics/Ampere%27s_Law

==References==

Chabay, Sherwood. (2015). Matter and Interactions (4th ed., Vol. 2). Raleigh, North Carolina: Wiley. page 887.

[[Category:Maxwell's Equations]]

Magnetic Field of a Long Thick Wire Using Ampere's Law

2016-11-28T03:30:07Z

Jeong-Eun Moon: /* Middling */

'''Claimed by Jeong-Eun Moon (Fall 2016)'''

==The Main Idea==

This section explains how to find the magnetic field near a long thick wire using Ampere's Law. Finding the magnetic field using Ampere's Law is very simple compared to finding it using the Biot-Savart law.

===Formula for Ampere's Law===

Here is the proof:

[[File:Ampere's_Law_Proof.png]]

[[File:Ampere's_Law.png]]

==How to Find Magnetic Field of A Long Thick Wire==

[[File:Wire.png]]

To find the magnetic field <math>B</math> at a distance <math>r</math> from the center of the long wire apply Ampere's Law. By the symmetry of the wire <math>B</math> will always be constant and tangential to the circular path at every point around the wire.

The magnetic field is everywhere parallel to the path for a circular path centered on wire. The direction of magnetic field can be determined by the right hand rule. The magnetic field's direction is perpendicular to the wire and is in the direction the fingers curl if you wrap the wire around. The direction of current is where your thumb points to.

[[File:Ampere's_Picture.png]]

The path integral <math>{{\oint}d\vec{l}}</math> in this situation is equal to the circumference of the circular path around the wire. This is equal to <math>2πr</math>.

Using the formula above and plugging in <math>{d\vec{l}}</math> we have: <math>{B(2πr) = μ_0I}</math>. To solve for <math>B</math> divide both sides by <math>2πr</math>.

This results in the equation: <math>{B = \frac{μ_0I}{2πr}}</math> which is equal to <math>{\frac{μ_02I}{4πr}}</math>. This is the equation for the magnetic field of a long thick wire that is found using the Biot-Savart law.

==Examples==

===Simple===

What is the magnetic field at a point 0.03 m away from a wire that has a current of 7 amperes?

Solution: <math>I=7</math> and <math>r=0.03</math>. So inserting this into the formula gives <math>{B = \frac{μ_0(7)}{2π(0.03)}}</math>. This results in 4.67e-5 T

===Middling===

A long straight wire suspended in the air carries a conventional current of 7.4 amperes in the -x direction as shown (the wire runs along the x-axis). At a particular instant an electron at location < 0, -0.004, 0 > m has velocity < -3.5 e5, -4.2 e5, 0 > m/s. What is the magnetic field due to the wire at the location of the electron?

Solution: <math>{B = \frac{μ_0(7.4)}{2π(0.004)}}</math> and use right hand rule. The answer is < 0, 0, 3.7e-4 > T.

===Hard===

Using the same values as in the middling problem calculate the magnetic force on the electron due to the wire.

Solution: The formula for magnetic force is <math>qv×B</math>. So by using the value for B computed above and calculating the cross product between B and qv we find that this equals < 2.49e-17, -2.07e-17, 0 > N. Don't forget that because this is the force on an electron q is negative.

== See also ==

See the page on Ampere's Law for a more in depth look at the law itself: [[Ampere's Law]]

For more applications of Ampere's Law see: [[Magnetic Field of a Toroid Using Ampere's Law]] and [[Magnetic Field of Coaxial Cable Using Ampere's Law]]

===Further reading===

Chabay, Sherwood. (2015). Matter and Interactions (4th ed., Vol. 2). Raleigh, North Carolina: Wiley.

===External links===

http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/magcur.html

https://www.khanacademy.org/science/physics/magnetic-forces-and-magnetic-fields/magnetic-field-current-carrying-wire/v/magnetism-6-magnetic-field-due-to-current

==References==

Chabay, Sherwood. (2015). Matter and Interactions (4th ed., Vol. 2). Raleigh, North Carolina: Wiley. page 887.

[[Category:Maxwell's Equations]]

Magnetic Field of a Long Thick Wire Using Ampere's Law

2016-11-26T03:49:45Z

Jeong-Eun Moon: /* Examples */

'''Claimed by Jeong-Eun Moon (Fall 2016)'''

==The Main Idea==

This section explains how to find the magnetic field near a long thick wire using Ampere's Law. Finding the magnetic field using Ampere's Law is very simple compared to finding it using the Biot-Savart law.

===Formula for Ampere's Law===

Here is the proof:

[[File:Ampere's_Law_Proof.png]]

[[File:Ampere's_Law.png]]

==How to Find Magnetic Field of A Long Thick Wire==

[[File:Wire.png]]

To find the magnetic field <math>B</math> at a distance <math>r</math> from the center of the long wire apply Ampere's Law. By the symmetry of the wire <math>B</math> will always be constant and tangential to the circular path at every point around the wire.

The magnetic field is everywhere parallel to the path for a circular path centered on wire. The direction of magnetic field can be determined by the right hand rule. The magnetic field's direction is perpendicular to the wire and is in the direction the fingers curl if you wrap the wire around. The direction of current is where your thumb points to.

[[File:Ampere's_Picture.png]]

The path integral <math>{{\oint}d\vec{l}}</math> in this situation is equal to the circumference of the circular path around the wire. This is equal to <math>2πr</math>.

Using the formula above and plugging in <math>{d\vec{l}}</math> we have: <math>{B(2πr) = μ_0I}</math>. To solve for <math>B</math> divide both sides by <math>2πr</math>.

This results in the equation: <math>{B = \frac{μ_0I}{2πr}}</math> which is equal to <math>{\frac{μ_02I}{4πr}}</math>. This is the equation for the magnetic field of a long thick wire that is found using the Biot-Savart law.

==Examples==

===Simple===

What is the magnetic field at a point 0.03 m away from a wire that has a current of 7 amperes?

Solution: <math>I=7</math> and <math>r=0.03</math>. So inserting this into the formula gives <math>{B = \frac{μ_0(7)}{2π(0.03)}}</math>. This results in 4.67e-5 T

===Middling===

A long straight wire suspended in the air carries a conventional current of 7.4 amperes in the -x direction as shown (the wire runs along the x-axis). At a particular instant an electron at location < 0, -0.004, 0 > m has velocity < -3.5 e5, -4.2 e5, 0 > m/s. What is the magnetic field due to the wire at the location of the electron?

Solution: <math>{B = \frac{μ_0(7.4)}{2π(0.004)}}</math>
< 0, 0, 3.7e-4 > T.

===Hard===

Using the same values as in the middling problem calculate the magnetic force on the electron due to the wire.

Solution: The formula for magnetic force is <math>qv×B</math>. So by using the value for B computed above and calculating the cross product between B and qv we find that this equals < 2.49e-17, -2.07e-17, 0 > N. Don't forget that because this is the force on an electron q is negative.

== See also ==

See the page on Ampere's Law for a more in depth look at the law itself: [[Ampere's Law]]

For more applications of Ampere's Law see: [[Magnetic Field of a Toroid Using Ampere's Law]] and [[Magnetic Field of Coaxial Cable Using Ampere's Law]]

===Further reading===

Chabay, Sherwood. (2015). Matter and Interactions (4th ed., Vol. 2). Raleigh, North Carolina: Wiley.

===External links===

http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/magcur.html

https://www.khanacademy.org/science/physics/magnetic-forces-and-magnetic-fields/magnetic-field-current-carrying-wire/v/magnetism-6-magnetic-field-due-to-current

==References==

Chabay, Sherwood. (2015). Matter and Interactions (4th ed., Vol. 2). Raleigh, North Carolina: Wiley. page 887.

[[Category:Maxwell's Equations]]

Magnetic Field of a Long Thick Wire Using Ampere's Law

2016-11-26T03:48:30Z

Jeong-Eun Moon: /* Examples */

'''Claimed by Jeong-Eun Moon (Fall 2016)'''

==The Main Idea==

This section explains how to find the magnetic field near a long thick wire using Ampere's Law. Finding the magnetic field using Ampere's Law is very simple compared to finding it using the Biot-Savart law.

===Formula for Ampere's Law===

Here is the proof:

[[File:Ampere's_Law_Proof.png]]

[[File:Ampere's_Law.png]]

==How to Find Magnetic Field of A Long Thick Wire==

[[File:Wire.png]]

To find the magnetic field <math>B</math> at a distance <math>r</math> from the center of the long wire apply Ampere's Law. By the symmetry of the wire <math>B</math> will always be constant and tangential to the circular path at every point around the wire.

The magnetic field is everywhere parallel to the path for a circular path centered on wire. The direction of magnetic field can be determined by the right hand rule. The magnetic field's direction is perpendicular to the wire and is in the direction the fingers curl if you wrap the wire around. The direction of current is where your thumb points to.

[[File:Ampere's_Picture.png]]

The path integral <math>{{\oint}d\vec{l}}</math> in this situation is equal to the circumference of the circular path around the wire. This is equal to <math>2πr</math>.

Using the formula above and plugging in <math>{d\vec{l}}</math> we have: <math>{B(2πr) = μ_0I}</math>. To solve for <math>B</math> divide both sides by <math>2πr</math>.

This results in the equation: <math>{B = \frac{μ_0I}{2πr}}</math> which is equal to <math>{\frac{μ_02I}{4πr}}</math>. This is the equation for the magnetic field of a long thick wire that is found using the Biot-Savart law.

==Examples==

===Simple===

What is the magnetic field at a point 0.03 m away from a wire that has a current of 7 amperes?

Solution: <math>I=7</math> and <math>r=0.03</math>. So inserting this into the formula gives <math>{B = \frac{μ_0(7)}{2π(0.03)}}</math>. This results in 4.67e-5 T

===Middling===

A long straight wire suspended in the air carries a conventional current of 7.4 amperes in the -x direction as shown (the wire runs along the x-axis). At a particular instant an electron at location < 0, -0.004, 0 > m has velocity < -3.5 e5, -4.2 e5, 0 > m/s. What is the magnetic field due to the wire at the location of the electron?

Solution: <math>{B = \frac{μ_0(7)}{2π(0.03)}}</math>
< 0, 0, 3.7e-4 > T.

===Hard===

Using the same values as in the middling problem calculate the magnetic force on the electron due to the wire.

Solution: The formula for magnetic force is <math>qv×B</math>. So by using the value for B computed above and calculating the cross product between B and qv we find that this equals < 2.49e-17, -2.07e-17, 0 > N. Don't forget that because this is the force on an electron q is negative.

== See also ==

See the page on Ampere's Law for a more in depth look at the law itself: [[Ampere's Law]]

For more applications of Ampere's Law see: [[Magnetic Field of a Toroid Using Ampere's Law]] and [[Magnetic Field of Coaxial Cable Using Ampere's Law]]

===Further reading===

Chabay, Sherwood. (2015). Matter and Interactions (4th ed., Vol. 2). Raleigh, North Carolina: Wiley.

===External links===

http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/magcur.html

https://www.khanacademy.org/science/physics/magnetic-forces-and-magnetic-fields/magnetic-field-current-carrying-wire/v/magnetism-6-magnetic-field-due-to-current

==References==

Chabay, Sherwood. (2015). Matter and Interactions (4th ed., Vol. 2). Raleigh, North Carolina: Wiley. page 887.

[[Category:Maxwell's Equations]]

Magnetic Field of a Long Thick Wire Using Ampere's Law

2016-11-26T03:47:34Z

Jeong-Eun Moon: /* How to Find Magnetic Field of A Long Thick Wire */

'''Claimed by Jeong-Eun Moon (Fall 2016)'''

==The Main Idea==

This section explains how to find the magnetic field near a long thick wire using Ampere's Law. Finding the magnetic field using Ampere's Law is very simple compared to finding it using the Biot-Savart law.

===Formula for Ampere's Law===

Here is the proof:

[[File:Ampere's_Law_Proof.png]]

[[File:Ampere's_Law.png]]

==How to Find Magnetic Field of A Long Thick Wire==

[[File:Wire.png]]

To find the magnetic field <math>B</math> at a distance <math>r</math> from the center of the long wire apply Ampere's Law. By the symmetry of the wire <math>B</math> will always be constant and tangential to the circular path at every point around the wire.

The magnetic field is everywhere parallel to the path for a circular path centered on wire. The direction of magnetic field can be determined by the right hand rule. The magnetic field's direction is perpendicular to the wire and is in the direction the fingers curl if you wrap the wire around. The direction of current is where your thumb points to.

[[File:Ampere's_Picture.png]]

The path integral <math>{{\oint}d\vec{l}}</math> in this situation is equal to the circumference of the circular path around the wire. This is equal to <math>2πr</math>.

Using the formula above and plugging in <math>{d\vec{l}}</math> we have: <math>{B(2πr) = μ_0I}</math>. To solve for <math>B</math> divide both sides by <math>2πr</math>.

This results in the equation: <math>{B = \frac{μ_0I}{2πr}}</math> which is equal to <math>{\frac{μ_02I}{4πr}}</math>. This is the equation for the magnetic field of a long thick wire that is found using the Biot-Savart law.

==Examples==

===Simple===

What is the magnetic field at a point 0.03 m away from a wire that has a current of 7 amperes?

Solution: <math>I=7</math> and <math>r=0.03</math>. So inserting this into the formula gives <math>{B = \frac{μ_0(7)}{2π(0.03)}}</math>. This results in 4.67e-5 T

===Middling===

A long straight wire suspended in the air carries a conventional current of 7.4 amperes in the -x direction as shown (the wire runs along the x-axis). At a particular instant an electron at location < 0, -0.004, 0 > m has velocity < -3.5 e5, -4.2 e5, 0 > m/s. What is the magnetic field due to the wire at the location of the electron?

Solution: Using the same formula as above and also implementing the right hand rule this equates to < 0, 0, 3.7e-4 > T.

===Hard===

Using the same values as in the middling problem calculate the magnetic force on the electron due to the wire.

Solution: The formula for magnetic force is <math>qv×B</math>. So by using the value for B computed above and calculating the cross product between B and qv we find that this equals < 2.49e-17, -2.07e-17, 0 > N. Don't forget that because this is the force on an electron q is negative.

== See also ==

See the page on Ampere's Law for a more in depth look at the law itself: [[Ampere's Law]]

For more applications of Ampere's Law see: [[Magnetic Field of a Toroid Using Ampere's Law]] and [[Magnetic Field of Coaxial Cable Using Ampere's Law]]

===Further reading===

Chabay, Sherwood. (2015). Matter and Interactions (4th ed., Vol. 2). Raleigh, North Carolina: Wiley.

===External links===

http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/magcur.html

https://www.khanacademy.org/science/physics/magnetic-forces-and-magnetic-fields/magnetic-field-current-carrying-wire/v/magnetism-6-magnetic-field-due-to-current

==References==

Chabay, Sherwood. (2015). Matter and Interactions (4th ed., Vol. 2). Raleigh, North Carolina: Wiley. page 887.

[[Category:Maxwell's Equations]]

Magnetic Field of a Long Thick Wire Using Ampere's Law

2016-11-26T03:39:16Z

Jeong-Eun Moon: /* How to Find Magnetic Field of A Long Thick Wire */

'''Claimed by Jeong-Eun Moon (Fall 2016)'''

==The Main Idea==

This section explains how to find the magnetic field near a long thick wire using Ampere's Law. Finding the magnetic field using Ampere's Law is very simple compared to finding it using the Biot-Savart law.

===Formula for Ampere's Law===

Here is the proof:

[[File:Ampere's_Law_Proof.png]]

[[File:Ampere's_Law.png]]

==How to Find Magnetic Field of A Long Thick Wire==

[[File:Wire.png]]

To find the magnetic field <math>B</math> at a distance <math>r</math> from the center of the long wire apply Ampere's Law. By the symmetry of the wire <math>B</math> will always be constant and tangential to the circular path at every point around the wire.

The magnetic field is everywhere parallel to the path for a circular path centered on wire.

[[File:Ampere's_Picture.png]]

The path integral <math>{{\oint}d\vec{l}}</math> in this situation is equal to the circumference of the circular path around the wire. This is equal to <math>2πr</math>.

Using the formula above and plugging in <math>{d\vec{l}}</math> we have: <math>{B(2πr) = μ_0I}</math>. To solve for <math>B</math> divide both sides by <math>2πr</math>.

This results in the equation: <math>{B = \frac{μ_0I}{2πr}}</math> which is equal to <math>{\frac{μ_02I}{4πr}}</math>. This is the equation for the magnetic field of a long thick wire that is found using the Biot-Savart law.

==Examples==

===Simple===

What is the magnetic field at a point 0.03 m away from a wire that has a current of 7 amperes?

Solution: <math>I=7</math> and <math>r=0.03</math>. So inserting this into the formula gives <math>{B = \frac{μ_0(7)}{2π(0.03)}}</math>. This results in 4.67e-5 T

===Middling===

A long straight wire suspended in the air carries a conventional current of 7.4 amperes in the -x direction as shown (the wire runs along the x-axis). At a particular instant an electron at location < 0, -0.004, 0 > m has velocity < -3.5 e5, -4.2 e5, 0 > m/s. What is the magnetic field due to the wire at the location of the electron?

Solution: Using the same formula as above and also implementing the right hand rule this equates to < 0, 0, 3.7e-4 > T.

===Hard===

Using the same values as in the middling problem calculate the magnetic force on the electron due to the wire.

Solution: The formula for magnetic force is <math>qv×B</math>. So by using the value for B computed above and calculating the cross product between B and qv we find that this equals < 2.49e-17, -2.07e-17, 0 > N. Don't forget that because this is the force on an electron q is negative.

== See also ==

See the page on Ampere's Law for a more in depth look at the law itself: [[Ampere's Law]]

For more applications of Ampere's Law see: [[Magnetic Field of a Toroid Using Ampere's Law]] and [[Magnetic Field of Coaxial Cable Using Ampere's Law]]

===Further reading===

Chabay, Sherwood. (2015). Matter and Interactions (4th ed., Vol. 2). Raleigh, North Carolina: Wiley.

===External links===

http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/magcur.html

https://www.khanacademy.org/science/physics/magnetic-forces-and-magnetic-fields/magnetic-field-current-carrying-wire/v/magnetism-6-magnetic-field-due-to-current

==References==

Chabay, Sherwood. (2015). Matter and Interactions (4th ed., Vol. 2). Raleigh, North Carolina: Wiley. page 887.

[[Category:Maxwell's Equations]]

File:Ampere's Picture.png

2016-11-26T03:38:50Z

Jeong-Eun Moon:

Magnetic Field of a Long Thick Wire Using Ampere's Law

2016-11-26T03:38:02Z

Jeong-Eun Moon: /* How to Find Magnetic Field of A Long Thick Wire */

'''Claimed by Jeong-Eun Moon (Fall 2016)'''

==The Main Idea==

This section explains how to find the magnetic field near a long thick wire using Ampere's Law. Finding the magnetic field using Ampere's Law is very simple compared to finding it using the Biot-Savart law.

===Formula for Ampere's Law===

Here is the proof:

[[File:Ampere's_Law_Proof.png]]

[[File:Ampere's_Law.png]]

==How to Find Magnetic Field of A Long Thick Wire==

[[File:Wire.png]]

To find the magnetic field <math>B</math> at a distance <math>r</math> from the center of the long wire apply Ampere's Law. By the symmetry of the wire <math>B</math> will always be constant and tangential to the circular path at every point around the wire.

The magnetic field is everywhere parallel to the path for a circular path centered on wire.

The path integral <math>{{\oint}d\vec{l}}</math> in this situation is equal to the circumference of the circular path around the wire. This is equal to <math>2πr</math>.

Using the formula above and plugging in <math>{d\vec{l}}</math> we have: <math>{B(2πr) = μ_0I}</math>. To solve for <math>B</math> divide both sides by <math>2πr</math>.

This results in the equation: <math>{B = \frac{μ_0I}{2πr}}</math> which is equal to <math>{\frac{μ_02I}{4πr}}</math>. This is the equation for the magnetic field of a long thick wire that is found using the Biot-Savart law.

==Examples==

===Simple===

What is the magnetic field at a point 0.03 m away from a wire that has a current of 7 amperes?

Solution: <math>I=7</math> and <math>r=0.03</math>. So inserting this into the formula gives <math>{B = \frac{μ_0(7)}{2π(0.03)}}</math>. This results in 4.67e-5 T

===Middling===

A long straight wire suspended in the air carries a conventional current of 7.4 amperes in the -x direction as shown (the wire runs along the x-axis). At a particular instant an electron at location < 0, -0.004, 0 > m has velocity < -3.5 e5, -4.2 e5, 0 > m/s. What is the magnetic field due to the wire at the location of the electron?

Solution: Using the same formula as above and also implementing the right hand rule this equates to < 0, 0, 3.7e-4 > T.

===Hard===

Using the same values as in the middling problem calculate the magnetic force on the electron due to the wire.

Solution: The formula for magnetic force is <math>qv×B</math>. So by using the value for B computed above and calculating the cross product between B and qv we find that this equals < 2.49e-17, -2.07e-17, 0 > N. Don't forget that because this is the force on an electron q is negative.

== See also ==

See the page on Ampere's Law for a more in depth look at the law itself: [[Ampere's Law]]

For more applications of Ampere's Law see: [[Magnetic Field of a Toroid Using Ampere's Law]] and [[Magnetic Field of Coaxial Cable Using Ampere's Law]]

===Further reading===

Chabay, Sherwood. (2015). Matter and Interactions (4th ed., Vol. 2). Raleigh, North Carolina: Wiley.

===External links===

http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/magcur.html

https://www.khanacademy.org/science/physics/magnetic-forces-and-magnetic-fields/magnetic-field-current-carrying-wire/v/magnetism-6-magnetic-field-due-to-current

==References==

Chabay, Sherwood. (2015). Matter and Interactions (4th ed., Vol. 2). Raleigh, North Carolina: Wiley. page 887.

[[Category:Maxwell's Equations]]

Magnetic Field of a Long Thick Wire Using Ampere's Law

2016-11-26T03:16:53Z

Jeong-Eun Moon: /* Formula for Ampere's Law */

'''Claimed by Jeong-Eun Moon (Fall 2016)'''

==The Main Idea==

This section explains how to find the magnetic field near a long thick wire using Ampere's Law. Finding the magnetic field using Ampere's Law is very simple compared to finding it using the Biot-Savart law.

===Formula for Ampere's Law===

Here is the proof:

[[File:Ampere's_Law_Proof.png]]

[[File:Ampere's_Law.png]]

==How to Find Magnetic Field of A Long Thick Wire==

[[File:Wire.png]]

To find the magnetic field <math>B</math> at a distance <math>r</math> from the center of the long wire apply Ampere's Law. By the symmetry of the wire <math>B</math> will always be constant and tangential to the circular path at every point around the wire.

The path integral <math>{{\oint}d\vec{l}}</math> in this situation is equal to the circumference of the circular path around the wire. This is equal to <math>2πr</math>.

Using the formula above and plugging in <math>{d\vec{l}}</math> we have: <math>{B(2πr) = μ_0I}</math>. To solve for <math>B</math> divide both sides by <math>2πr</math>.

This results in the equation: <math>{B = \frac{μ_0I}{2πr}}</math> which is equal to <math>{\frac{μ_02I}{4πr}}</math>. This is the equation for the magnetic field of a long thick wire that is found using the Biot-Savart law.

==Examples==

===Simple===

What is the magnetic field at a point 0.03 m away from a wire that has a current of 7 amperes?

Solution: <math>I=7</math> and <math>r=0.03</math>. So inserting this into the formula gives <math>{B = \frac{μ_0(7)}{2π(0.03)}}</math>. This results in 4.67e-5 T

===Middling===

A long straight wire suspended in the air carries a conventional current of 7.4 amperes in the -x direction as shown (the wire runs along the x-axis). At a particular instant an electron at location < 0, -0.004, 0 > m has velocity < -3.5 e5, -4.2 e5, 0 > m/s. What is the magnetic field due to the wire at the location of the electron?

Solution: Using the same formula as above and also implementing the right hand rule this equates to < 0, 0, 3.7e-4 > T.

===Hard===

Using the same values as in the middling problem calculate the magnetic force on the electron due to the wire.

Solution: The formula for magnetic force is <math>qv×B</math>. So by using the value for B computed above and calculating the cross product between B and qv we find that this equals < 2.49e-17, -2.07e-17, 0 > N. Don't forget that because this is the force on an electron q is negative.

== See also ==

See the page on Ampere's Law for a more in depth look at the law itself: [[Ampere's Law]]

For more applications of Ampere's Law see: [[Magnetic Field of a Toroid Using Ampere's Law]] and [[Magnetic Field of Coaxial Cable Using Ampere's Law]]

===Further reading===

Chabay, Sherwood. (2015). Matter and Interactions (4th ed., Vol. 2). Raleigh, North Carolina: Wiley.

===External links===

http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/magcur.html

https://www.khanacademy.org/science/physics/magnetic-forces-and-magnetic-fields/magnetic-field-current-carrying-wire/v/magnetism-6-magnetic-field-due-to-current

==References==

Chabay, Sherwood. (2015). Matter and Interactions (4th ed., Vol. 2). Raleigh, North Carolina: Wiley. page 887.

[[Category:Maxwell's Equations]]

Magnetic Field of a Long Thick Wire Using Ampere's Law

2016-11-26T03:16:39Z

Jeong-Eun Moon: /* Formula for Ampere's Law */

'''Claimed by Jeong-Eun Moon (Fall 2016)'''

==The Main Idea==

This section explains how to find the magnetic field near a long thick wire using Ampere's Law. Finding the magnetic field using Ampere's Law is very simple compared to finding it using the Biot-Savart law.

===Formula for Ampere's Law===

Here is the proof:
[[File:Ampere's_Law_Proof.png]]

[[File:Ampere's_Law.png]]

==How to Find Magnetic Field of A Long Thick Wire==

[[File:Wire.png]]

To find the magnetic field <math>B</math> at a distance <math>r</math> from the center of the long wire apply Ampere's Law. By the symmetry of the wire <math>B</math> will always be constant and tangential to the circular path at every point around the wire.

The path integral <math>{{\oint}d\vec{l}}</math> in this situation is equal to the circumference of the circular path around the wire. This is equal to <math>2πr</math>.

Using the formula above and plugging in <math>{d\vec{l}}</math> we have: <math>{B(2πr) = μ_0I}</math>. To solve for <math>B</math> divide both sides by <math>2πr</math>.

This results in the equation: <math>{B = \frac{μ_0I}{2πr}}</math> which is equal to <math>{\frac{μ_02I}{4πr}}</math>. This is the equation for the magnetic field of a long thick wire that is found using the Biot-Savart law.

==Examples==

===Simple===

What is the magnetic field at a point 0.03 m away from a wire that has a current of 7 amperes?

Solution: <math>I=7</math> and <math>r=0.03</math>. So inserting this into the formula gives <math>{B = \frac{μ_0(7)}{2π(0.03)}}</math>. This results in 4.67e-5 T

===Middling===

A long straight wire suspended in the air carries a conventional current of 7.4 amperes in the -x direction as shown (the wire runs along the x-axis). At a particular instant an electron at location < 0, -0.004, 0 > m has velocity < -3.5 e5, -4.2 e5, 0 > m/s. What is the magnetic field due to the wire at the location of the electron?

Solution: Using the same formula as above and also implementing the right hand rule this equates to < 0, 0, 3.7e-4 > T.

===Hard===

Using the same values as in the middling problem calculate the magnetic force on the electron due to the wire.

Solution: The formula for magnetic force is <math>qv×B</math>. So by using the value for B computed above and calculating the cross product between B and qv we find that this equals < 2.49e-17, -2.07e-17, 0 > N. Don't forget that because this is the force on an electron q is negative.

== See also ==

See the page on Ampere's Law for a more in depth look at the law itself: [[Ampere's Law]]

For more applications of Ampere's Law see: [[Magnetic Field of a Toroid Using Ampere's Law]] and [[Magnetic Field of Coaxial Cable Using Ampere's Law]]

===Further reading===

Chabay, Sherwood. (2015). Matter and Interactions (4th ed., Vol. 2). Raleigh, North Carolina: Wiley.

===External links===

http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/magcur.html

https://www.khanacademy.org/science/physics/magnetic-forces-and-magnetic-fields/magnetic-field-current-carrying-wire/v/magnetism-6-magnetic-field-due-to-current

==References==

Chabay, Sherwood. (2015). Matter and Interactions (4th ed., Vol. 2). Raleigh, North Carolina: Wiley. page 887.

[[Category:Maxwell's Equations]]

File:Ampere's Law Proof.png

2016-11-26T03:15:47Z

Jeong-Eun Moon:

Magnetic Field of a Long Thick Wire Using Ampere's Law

2016-11-26T03:13:40Z

Jeong-Eun Moon: /* Formula for Ampere's Law */

'''Claimed by Jeong-Eun Moon (Fall 2016)'''

==The Main Idea==

This section explains how to find the magnetic field near a long thick wire using Ampere's Law. Finding the magnetic field using Ampere's Law is very simple compared to finding it using the Biot-Savart law.

===Formula for Ampere's Law===

Here is the proof:

[[File:Ampere's_Law.png]]

==How to Find Magnetic Field of A Long Thick Wire==

[[File:Wire.png]]

To find the magnetic field <math>B</math> at a distance <math>r</math> from the center of the long wire apply Ampere's Law. By the symmetry of the wire <math>B</math> will always be constant and tangential to the circular path at every point around the wire.

The path integral <math>{{\oint}d\vec{l}}</math> in this situation is equal to the circumference of the circular path around the wire. This is equal to <math>2πr</math>.

Using the formula above and plugging in <math>{d\vec{l}}</math> we have: <math>{B(2πr) = μ_0I}</math>. To solve for <math>B</math> divide both sides by <math>2πr</math>.

This results in the equation: <math>{B = \frac{μ_0I}{2πr}}</math> which is equal to <math>{\frac{μ_02I}{4πr}}</math>. This is the equation for the magnetic field of a long thick wire that is found using the Biot-Savart law.

==Examples==

===Simple===

What is the magnetic field at a point 0.03 m away from a wire that has a current of 7 amperes?

Solution: <math>I=7</math> and <math>r=0.03</math>. So inserting this into the formula gives <math>{B = \frac{μ_0(7)}{2π(0.03)}}</math>. This results in 4.67e-5 T

===Middling===

A long straight wire suspended in the air carries a conventional current of 7.4 amperes in the -x direction as shown (the wire runs along the x-axis). At a particular instant an electron at location < 0, -0.004, 0 > m has velocity < -3.5 e5, -4.2 e5, 0 > m/s. What is the magnetic field due to the wire at the location of the electron?

Solution: Using the same formula as above and also implementing the right hand rule this equates to < 0, 0, 3.7e-4 > T.

===Hard===

Using the same values as in the middling problem calculate the magnetic force on the electron due to the wire.

Solution: The formula for magnetic force is <math>qv×B</math>. So by using the value for B computed above and calculating the cross product between B and qv we find that this equals < 2.49e-17, -2.07e-17, 0 > N. Don't forget that because this is the force on an electron q is negative.

== See also ==

See the page on Ampere's Law for a more in depth look at the law itself: [[Ampere's Law]]

For more applications of Ampere's Law see: [[Magnetic Field of a Toroid Using Ampere's Law]] and [[Magnetic Field of Coaxial Cable Using Ampere's Law]]

===Further reading===

Chabay, Sherwood. (2015). Matter and Interactions (4th ed., Vol. 2). Raleigh, North Carolina: Wiley.

===External links===

http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/magcur.html

https://www.khanacademy.org/science/physics/magnetic-forces-and-magnetic-fields/magnetic-field-current-carrying-wire/v/magnetism-6-magnetic-field-due-to-current

==References==

Chabay, Sherwood. (2015). Matter and Interactions (4th ed., Vol. 2). Raleigh, North Carolina: Wiley. page 887.

[[Category:Maxwell's Equations]]

Magnetic Field of a Long Thick Wire Using Ampere's Law

2016-11-26T03:13:24Z

Jeong-Eun Moon: /* Formula for Ampere's Law */

'''Claimed by Jeong-Eun Moon (Fall 2016)'''

==The Main Idea==

This section explains how to find the magnetic field near a long thick wire using Ampere's Law. Finding the magnetic field using Ampere's Law is very simple compared to finding it using the Biot-Savart law.

===Formula for Ampere's Law===

Here is the proof:
[[File:proof.png]]
[[File:Ampere's_Law.png]]

==How to Find Magnetic Field of A Long Thick Wire==

[[File:Wire.png]]

To find the magnetic field <math>B</math> at a distance <math>r</math> from the center of the long wire apply Ampere's Law. By the symmetry of the wire <math>B</math> will always be constant and tangential to the circular path at every point around the wire.

The path integral <math>{{\oint}d\vec{l}}</math> in this situation is equal to the circumference of the circular path around the wire. This is equal to <math>2πr</math>.

Using the formula above and plugging in <math>{d\vec{l}}</math> we have: <math>{B(2πr) = μ_0I}</math>. To solve for <math>B</math> divide both sides by <math>2πr</math>.

This results in the equation: <math>{B = \frac{μ_0I}{2πr}}</math> which is equal to <math>{\frac{μ_02I}{4πr}}</math>. This is the equation for the magnetic field of a long thick wire that is found using the Biot-Savart law.

==Examples==

===Simple===

What is the magnetic field at a point 0.03 m away from a wire that has a current of 7 amperes?

Solution: <math>I=7</math> and <math>r=0.03</math>. So inserting this into the formula gives <math>{B = \frac{μ_0(7)}{2π(0.03)}}</math>. This results in 4.67e-5 T

===Middling===

A long straight wire suspended in the air carries a conventional current of 7.4 amperes in the -x direction as shown (the wire runs along the x-axis). At a particular instant an electron at location < 0, -0.004, 0 > m has velocity < -3.5 e5, -4.2 e5, 0 > m/s. What is the magnetic field due to the wire at the location of the electron?

Solution: Using the same formula as above and also implementing the right hand rule this equates to < 0, 0, 3.7e-4 > T.

===Hard===

Using the same values as in the middling problem calculate the magnetic force on the electron due to the wire.

Solution: The formula for magnetic force is <math>qv×B</math>. So by using the value for B computed above and calculating the cross product between B and qv we find that this equals < 2.49e-17, -2.07e-17, 0 > N. Don't forget that because this is the force on an electron q is negative.

== See also ==

See the page on Ampere's Law for a more in depth look at the law itself: [[Ampere's Law]]

For more applications of Ampere's Law see: [[Magnetic Field of a Toroid Using Ampere's Law]] and [[Magnetic Field of Coaxial Cable Using Ampere's Law]]

===Further reading===

Chabay, Sherwood. (2015). Matter and Interactions (4th ed., Vol. 2). Raleigh, North Carolina: Wiley.

===External links===

http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/magcur.html

https://www.khanacademy.org/science/physics/magnetic-forces-and-magnetic-fields/magnetic-field-current-carrying-wire/v/magnetism-6-magnetic-field-due-to-current

==References==

Chabay, Sherwood. (2015). Matter and Interactions (4th ed., Vol. 2). Raleigh, North Carolina: Wiley. page 887.

[[Category:Maxwell's Equations]]

Magnetic Field of a Long Thick Wire Using Ampere's Law

2016-11-26T03:10:39Z

Jeong-Eun Moon: /* Formula for Ampere's Law */

'''Claimed by Jeong-Eun Moon (Fall 2016)'''

==The Main Idea==

This section explains how to find the magnetic field near a long thick wire using Ampere's Law. Finding the magnetic field using Ampere's Law is very simple compared to finding it using the Biot-Savart law.

===Formula for Ampere's Law===

Here is the proof:
/Users/jeongeunmoon/Desktop/proof.png

[[File:Ampere's_Law.png]]

==How to Find Magnetic Field of A Long Thick Wire==

[[File:Wire.png]]

To find the magnetic field <math>B</math> at a distance <math>r</math> from the center of the long wire apply Ampere's Law. By the symmetry of the wire <math>B</math> will always be constant and tangential to the circular path at every point around the wire.

The path integral <math>{{\oint}d\vec{l}}</math> in this situation is equal to the circumference of the circular path around the wire. This is equal to <math>2πr</math>.

Using the formula above and plugging in <math>{d\vec{l}}</math> we have: <math>{B(2πr) = μ_0I}</math>. To solve for <math>B</math> divide both sides by <math>2πr</math>.

This results in the equation: <math>{B = \frac{μ_0I}{2πr}}</math> which is equal to <math>{\frac{μ_02I}{4πr}}</math>. This is the equation for the magnetic field of a long thick wire that is found using the Biot-Savart law.

==Examples==

===Simple===

What is the magnetic field at a point 0.03 m away from a wire that has a current of 7 amperes?

Solution: <math>I=7</math> and <math>r=0.03</math>. So inserting this into the formula gives <math>{B = \frac{μ_0(7)}{2π(0.03)}}</math>. This results in 4.67e-5 T

===Middling===

A long straight wire suspended in the air carries a conventional current of 7.4 amperes in the -x direction as shown (the wire runs along the x-axis). At a particular instant an electron at location < 0, -0.004, 0 > m has velocity < -3.5 e5, -4.2 e5, 0 > m/s. What is the magnetic field due to the wire at the location of the electron?

Solution: Using the same formula as above and also implementing the right hand rule this equates to < 0, 0, 3.7e-4 > T.

===Hard===

Using the same values as in the middling problem calculate the magnetic force on the electron due to the wire.

Solution: The formula for magnetic force is <math>qv×B</math>. So by using the value for B computed above and calculating the cross product between B and qv we find that this equals < 2.49e-17, -2.07e-17, 0 > N. Don't forget that because this is the force on an electron q is negative.

== See also ==

See the page on Ampere's Law for a more in depth look at the law itself: [[Ampere's Law]]

For more applications of Ampere's Law see: [[Magnetic Field of a Toroid Using Ampere's Law]] and [[Magnetic Field of Coaxial Cable Using Ampere's Law]]

===Further reading===

Chabay, Sherwood. (2015). Matter and Interactions (4th ed., Vol. 2). Raleigh, North Carolina: Wiley.

===External links===

http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/magcur.html

https://www.khanacademy.org/science/physics/magnetic-forces-and-magnetic-fields/magnetic-field-current-carrying-wire/v/magnetism-6-magnetic-field-due-to-current

==References==

Chabay, Sherwood. (2015). Matter and Interactions (4th ed., Vol. 2). Raleigh, North Carolina: Wiley. page 887.

[[Category:Maxwell's Equations]]

Conductivity

2016-11-14T22:17:07Z

Jeong-Eun Moon:

'''
''
Claimed by myoung65, Spring 2016

'''
==Definition==

Electrical Resistivity is a measure of how a given material opposes current flow. Low resistivity shows a material that allows the flow of current, whereas the opposite is true for high resistivity.
Electrical Conductivity is the reciprocal/inverse of Electrical Resistivity, in that it measures the ability of a given material to conduct electric current

==Symbols==

Electrical Resistivity is mainly represented by the Greek lower-case rho.
Electrical Conductivity is mainly represented by the Greek lower-case sigma, but is occasionally represented by a lower-case kappa, or gamma.

== SI Units ==

Electrical Resistivity is measured in Ohm-Metres.
Electrical Conductivity is measured in Siemens per Metre

== Classification of Materials by Conductivity ==

Materials with high Conductivity are known as conductors. ex. metals
Materials with low Conductivity are known as resistors. ex. vacuums, glass, etc.

== Semiconductors ==

Semiconductors are materials that have a conductivity in-between that of an insulator and a conductor. However, as temperature increases, unlike in most metals, the conductivity of semiconductors increases.

== Temperature Dependence ==

As temperature increases, the electrical resistivity of metals increases. This is a reason why when computers heat up, they tend to slow down. Some materials exhibit superconductivity at extremely low temperatures. Below a certain temperature, resistivity vanishes, such as Pb at 7.20 K.

==Equations==

'''Poulliet's Law'''

R=ρℓ/A

R = Electric Resistance
ρ = Electric Resistivity
ℓ = Length
A = Cross-Sectional Area

Poulliet's Law states that a given materials resistance will increase in length, while it will decrease with an increase in Area.

==Conductivity in Real Life==

Conductors are used to carry electricity, as well as electrical signals in circuits.
Complementary metal–oxide–semiconductors, or CMOS for short, are the foundational building block of gate based logic circuits, that make up the majority of all modern electronics. CMOS circuits are composed of a combination of p-type and n-type semiconductors. These semiconductors will change their conductivity, based on the applied voltage, allowing for logic of 0's and 1's, or low voltage and high voltage, to be transferred through logical circuits. This allows us to apply boolean logic to circuits, such as AND and OR logic, or even create an amalgamation of AND's and OR's to create electronics, such as multiplexors, switches, latches, registers, decoders, encoders, etc.

==History==

Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.

== See also ==

Are there related topics or categories in this wiki resource for the curious reader to explore? How does this topic fit into that context?

===Further reading===

Books, Articles or other print media on this topic

===External links===
[http://www.scientificamerican.com/article/bring-science-home-reaction-time/]

==References==

This section contains the the references you used while writing this page

[[Category:Which Category did you place this in?]]

Magnetic Field of a Long Thick Wire Using Ampere's Law

2016-11-14T22:16:19Z

Jeong-Eun Moon:

'''Claimed by Jeong-Eun Moon (Fall 2016)'''

==The Main Idea==

This section explains how to find the magnetic field near a long thick wire using Ampere's Law. Finding the magnetic field using Ampere's Law is very simple compared to finding it using the Biot-Savart law.

===Formula for Ampere's Law===
[[File:Ampere's_Law.png]]

==How to Find Magnetic Field of A Long Thick Wire==

[[File:Wire.png]]

To find the magnetic field <math>B</math> at a distance <math>r</math> from the center of the long wire apply Ampere's Law. By the symmetry of the wire <math>B</math> will always be constant and tangential to the circular path at every point around the wire.

The path integral <math>{{\oint}d\vec{l}}</math> in this situation is equal to the circumference of the circular path around the wire. This is equal to <math>2πr</math>.

Using the formula above and plugging in <math>{d\vec{l}}</math> we have: <math>{B(2πr) = μ_0I}</math>. To solve for <math>B</math> divide both sides by <math>2πr</math>.

This results in the equation: <math>{B = \frac{μ_0I}{2πr}}</math> which is equal to <math>{\frac{μ_02I}{4πr}}</math>. This is the equation for the magnetic field of a long thick wire that is found using the Biot-Savart law.

==Examples==

===Simple===

What is the magnetic field at a point 0.03 m away from a wire that has a current of 7 amperes?

Solution: <math>I=7</math> and <math>r=0.03</math>. So inserting this into the formula gives <math>{B = \frac{μ_0(7)}{2π(0.03)}}</math>. This results in 4.67e-5 T

===Middling===

A long straight wire suspended in the air carries a conventional current of 7.4 amperes in the -x direction as shown (the wire runs along the x-axis). At a particular instant an electron at location < 0, -0.004, 0 > m has velocity < -3.5 e5, -4.2 e5, 0 > m/s. What is the magnetic field due to the wire at the location of the electron?

Solution: Using the same formula as above and also implementing the right hand rule this equates to < 0, 0, 3.7e-4 > T.

===Hard===

Using the same values as in the middling problem calculate the magnetic force on the electron due to the wire.

Solution: The formula for magnetic force is <math>qv×B</math>. So by using the value for B computed above and calculating the cross product between B and qv we find that this equals < 2.49e-17, -2.07e-17, 0 > N. Don't forget that because this is the force on an electron q is negative.

== See also ==

See the page on Ampere's Law for a more in depth look at the law itself: [[Ampere's Law]]

For more applications of Ampere's Law see: [[Magnetic Field of a Toroid Using Ampere's Law]] and [[Magnetic Field of Coaxial Cable Using Ampere's Law]]

===Further reading===

Chabay, Sherwood. (2015). Matter and Interactions (4th ed., Vol. 2). Raleigh, North Carolina: Wiley.

===External links===

http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/magcur.html

https://www.khanacademy.org/science/physics/magnetic-forces-and-magnetic-fields/magnetic-field-current-carrying-wire/v/magnetism-6-magnetic-field-due-to-current

==References==

Chabay, Sherwood. (2015). Matter and Interactions (4th ed., Vol. 2). Raleigh, North Carolina: Wiley. page 887.

[[Category:Maxwell's Equations]]

Conductivity

2016-11-05T18:39:15Z

Jeong-Eun Moon:

'''
'''Claimed by Jeong-Eun Moon (jmoon79) - Fall 2016'''
Claimed by myoung65, Spring 2016

'''
==Definition==

Electrical Resistivity is a measure of how a given material opposes current flow. Low resistivity shows a material that allows the flow of current, whereas the opposite is true for high resistivity.
Electrical Conductivity is the reciprocal/inverse of Electrical Resistivity, in that it measures the ability of a given material to conduct electric current

==Symbols==

Electrical Resistivity is mainly represented by the Greek lower-case rho.
Electrical Conductivity is mainly represented by the Greek lower-case sigma, but is occasionally represented by a lower-case kappa, or gamma.

== SI Units ==

Electrical Resistivity is measured in Ohm-Metres.
Electrical Conductivity is measured in Siemens per Metre

== Classification of Materials by Conductivity ==

Materials with high Conductivity are known as conductors. ex. metals
Materials with low Conductivity are known as resistors. ex. vacuums, glass, etc.

== Semiconductors ==

Semiconductors are materials that have a conductivity in-between that of an insulator and a conductor. However, as temperature increases, unlike in most metals, the conductivity of semiconductors increases.

== Temperature Dependence ==

As temperature increases, the electrical resistivity of metals increases. This is a reason why when computers heat up, they tend to slow down. Some materials exhibit superconductivity at extremely low temperatures. Below a certain temperature, resistivity vanishes, such as Pb at 7.20 K.

==Equations==

'''Poulliet's Law'''

R=ρℓ/A

R = Electric Resistance
ρ = Electric Resistivity
ℓ = Length
A = Cross-Sectional Area

Poulliet's Law states that a given materials resistance will increase in length, while it will decrease with an increase in Area.

==Conductivity in Real Life==

Conductors are used to carry electricity, as well as electrical signals in circuits.
Complementary metal–oxide–semiconductors, or CMOS for short, are the foundational building block of gate based logic circuits, that make up the majority of all modern electronics. CMOS circuits are composed of a combination of p-type and n-type semiconductors. These semiconductors will change their conductivity, based on the applied voltage, allowing for logic of 0's and 1's, or low voltage and high voltage, to be transferred through logical circuits. This allows us to apply boolean logic to circuits, such as AND and OR logic, or even create an amalgamation of AND's and OR's to create electronics, such as multiplexors, switches, latches, registers, decoders, encoders, etc.

==History==

Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.

== See also ==

Are there related topics or categories in this wiki resource for the curious reader to explore? How does this topic fit into that context?

===Further reading===

Books, Articles or other print media on this topic

===External links===
[http://www.scientificamerican.com/article/bring-science-home-reaction-time/]

==References==

This section contains the the references you used while writing this page

[[Category:Which Category did you place this in?]]