Introduction to Ampere's Law
Ampere's Law establishes the relationship between an electric current and the magnetic field it produces. When a current flows through a wire, it generates a magnetic field surrounding the wire in circular loops. For a deeper understanding, see Understanding Ampere's Law and Its Application in Electromagnetism.
Magnetic Field Around a Long Straight Wire
- Right-hand rule determines the magnetic field direction: fingers curl around the wire in the direction of the magnetic field.
- Magnetic field (B) at distance R from a wire carrying current I is given by: [ B = \frac{\mu_0 I}{2 \pi R} ] where ( \mu_0 = 4 \pi \times 10^{-7} ) T·m/A is the permeability of free space.
- The magnetic field circulates around the wire, with field lines directed into or out of the page depending on the side of the wire. For more on magnetic fields near current-carrying wires, visit Calculating Magnetic Fields Around Straight Current-Carrying Wires.
Applying Ampere's Law to a Solenoid
- A solenoid is a coil of wire carrying current, producing a magnetic field strongest inside its center.
- Outside the solenoid, the magnetic field is negligible.
- Using a rectangular Amperian loop:
- The magnetic field is parallel and significant along the side inside the solenoid (segment AB).
- Perpendicular sides (BC and DA) contribute zero to the magnetic field line integral.
- The side outside the solenoid (CD) has a very weak magnetic field, approximated to zero.
Deriving Magnetic Field in a Solenoid
- Ampere's Law states: [ B \times L = \mu_0 \times I_{enc} ]
- Enclosed current ( I_{enc} = n \times I ), where:
- ( n ) = number of loops
- ( I ) = current per loop
- Convert to loops per unit length ( n' = \frac{n}{L} ), known as turn density.
- Rearranged formula for magnetic field inside solenoid: [ B = \mu_0 n' I ]
- This expression shows magnetic field strength depends on turn density and current.
- To broaden context on magnetic fields and currents, see Understanding Magnetism: Forces, Currents, and Magnetic Fields.
Example Calculation
- Solenoid parameters:
- Length ( L = 5 ) cm (0.05 m)
- Number of turns ( n = 1500 )
- Current ( I = 7 ) A
- Calculate turn density: [ n' = \frac{1500}{0.05} = 30000 \text{ turns/m} ]
- Calculate magnetic field: [ B = 4\pi \times 10^{-7} \times 30000 \times 7 = 0.26 \text{ Tesla} ]
- This shows a strong magnetic field at the solenoid's center.
Additional Resources
- For further study, explore the creator's physics playlist covering electricity, magnetism, and other common physics topics.
- Recommended follow-up video: "Electromagnetic Induction" for an in-depth understanding of related concepts.
Conclusion
Ampere's Law provides a powerful tool to calculate magnetic fields produced by currents in various configurations, including straight wires and solenoids. Understanding and applying this law is crucial for solving practical physics problems in electromagnetism. For a comprehensive overview of magnetic phenomena, you may also find Comprehensive Guide to Magnetism: Magnetic Fields, Forces, and Applications beneficial.
in this video I want to talk about amperes law amperes law highlights the relationship between the electric
current and magnetic field produced by such a current when you know the current that is enclosed in a magnetic loop so
let's use a long straight wire for example and let's say we have a current flowing in this wire now using the
right-hand rule this wire walk free a magnetic field that will flow in this direction on the right side is going to
go into the page and on the left side it comes out of the page now amperes law states that the sum of the magnetic
field the component the medic field times the distance that the magnetic field travels so the sum of these
products is equal to the permeability of free space times the current enclosed by the magnetic loop so notice that the
magnetic field is traveling in a circular path and the current enclosed by this magnetic Luc's so to speak is
the current that flows through this wire and so the enclosed current and simply I in this example and the left around the
circle is the circumference and the circumference of a circle is 2 pi R so we can replace out with 2 pi R so this
gives us the magnetic field of a long straight wire it's mu zero times I over 2 pi R so if we wish to calculate the
mimetic field at let's say point a all we need to know is the distance between the wire and pointing and the current
flowing through the wire and so it will give us this equation so the permeability of free space is 4 pi times
10 to the minus 7 so that's a constant that you need to know now let's use amperes law to divide the
formula that will help us to calculate the magnetic field of a solenoid so what is a solenoid it's basically just a coil
of wire so let's say we have a current flowing in this wider that current will create a magnetic field and the magnetic
field that passes through a solenoid its greatest at the center now it's going to go around in this general direction and
also here as well but we need to understand is that outside of the solenoid the mimetic field is very weak
but inside of the solenoid it's very very strong so let's say this is our loop of
interests let's use that as the magnetic loop and let's call this point a b c and d so let's calculate the sum of the
products of the mimetic field that is parallel to a segment of left which we'll call L so in this case that's
gonna be BL for a segment a B plus BL for segment BC and then B out for segment CD plus BL for segment da now
keep in mind the magnetic field is strongest at the center so notice that we have magnetic field that is parallel
to segment a B so we'll just call that B times L and in segment BC notice that it's perpendicular to this magnetic
field it's not parallel to it so that's going to be 0 and the same is true for segment da this segment is perpendicular
to this magnetic field so that's gonna be counted as zero and for CD even though there is a
magnetic field flown in this direction it's a very very small compared to the magnetic field that is at the centre of
the solenoid so we can say that this is approximately zero because it's very small and so it simplifies the
calculations so therefore using amperes law the sum of all the products of the magnetic field parallel to the segment
of each length is equal to MU 0 times 2 current enclosed so this quantity is simply B times L and that's equal to MU
0 times the enclosed current so what is the enclosed current in this example well if you have one loop well for
current of 10 amps won't do it the enclosed current is gonna be 10 amps now if you have 2 loops the enclosed current
doubles because the current is flowing around the loop twice now if we have three loops the enclosed current is
going to be 30 amps because we have a current of 10 amps flowing three times in a circle so a general expression for
the enclosed current is going to be the number of loops times the current flowing in the wire so therefore we have
this expression the L is equal to MU 0 unless replace this with n times I so and simply represents the number of
loops now to solve for the magnetic field B we need to divide both sides by L so on the left side this cancels now
what is the quantity n over L represents so this is equal to lowercase n where a low case n represents the number of
loops per meter so it's the number of loops per unit length so the magnetic field of a solenoid is going to be mu
zero times the number of loops per meter times the current and this gives you the minitek field inside of the solenoid and
so that's the general expression for it has to formally you need now let's work on an example problem a
five centimeter long solenoid has 1500 turns of wire and carries a current of 70 amps calculate the strength of the
magnetic field at the center of the solenoid so let's draw a picture so let's say we have a solenoid and it's 5
centimeters long and the number of turns is 1500 and we have a current of 7 amps flowing through it so I equals 7 so
calculate the strain for the magnetic field so we need to calculate the value of B so we know the formula its mu 0 and
I so first let's calculate n the number of turns per unit length so we have 1500 turns and the length of the solenoid is
5 centimeters which is 0.2 0.2 divided by point zero five it comes out to 30,000 turns per meter so that's the
value of lowercase n so now we could use that to calculate the strength of the magnetic field so B is going to equal mu
0 which is 4 pi times 10 to the minus 7 times n which is 30,000 multiplied by a current of 7 amps now let's go ahead and
get the answer for pi is basically twelve point five six six three seven if you take that and multiply by one times
10 to negative seven times 30 thousand times seven this will give you a magnetic field of point twenty six Tesla
and so that's how you can calculate the strength of the magnetic field at the centre of the solenoid
now for those of you who may want to find more of my physics videos you can check out my channel and I do have a
physics playlist which has electricity magnetism and all other stuff that you need in physics at least the common
topics that you might see in a typical high school or college level course so if you want access to those videos just
check out the playlist and you can find it at the end of the video in the last 20 seconds there should be an end screen
way if you click on it you can get access to that playlist so you can find more videos like this if you're
interested so thanks again for watching this video and have a good day by the way feel free to check out another video
that I've created right after magnetism it's a long video on YouTube but it's entitled
electromagnetic induction and I think you might find that video to be helpful as well
Ampere's Law states that the magnetic field around a closed loop is proportional to the electric current passing through the loop. Specifically, it quantifies that the magnetic field produced by a current-carrying wire forms circular loops around the wire, allowing calculation of field strength based on current and geometry.
You can use the right-hand rule: point the thumb of your right hand in the direction of the current flow, and your fingers will curl around the wire in the direction of the magnetic field lines. This helps visualize and determine whether the field points into or out of the page around the wire.
Inside a solenoid, the magnetic field is uniform and parallel to the axis. Using Ampere's Law and a rectangular Amperian loop, the magnetic field B inside is calculated as B = μ₀ × n' × I, where μ₀ is the permeability of free space, n' is the number of turns per unit length (turn density), and I is the current through each loop.
Outside the solenoid, the magnetic field lines spread out and mostly cancel due to the coil's geometry and the opposing magnetic fields produced by adjacent loops. As a result, the field strength outside is very weak compared to the strong, uniform field inside the solenoid, allowing it to be approximated as zero.
Yes. For a solenoid with 1500 turns, 5 cm length, and 7 A current: first calculate turn density n' = 1500 ÷ 0.05 m = 30000 turns/m. Then apply B = μ₀ × n' × I = 4π×10⁻⁷ × 30000 × 7 ≈ 0.26 Tesla. This shows a strong magnetic field inside the solenoid's center.
Increasing the number of turns per unit length (turn density) or the current flowing in the solenoid directly increases the magnetic field strength inside the coil. According to B = μ₀ × n' × I, the magnetic field is proportional to both the turn density and the current.
Ampere's Law and solenoid magnetic fields are foundational in designing electromagnets, transformers, inductors, electric motors, and MRI machines. Understanding these principles allows engineers to control magnetic field strengths and shapes crucial to electrical devices and magnetic sensing technologies.
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