Magnetic Fields Part 2: Sources of Magnetic Fields

Transcription

1 Magnetic Fields Part 2: Sources of Magnetic Fields Last modified: 08/01/2018

2 Contents Links What Causes a Magnetic Field? Moving Charges Right Hand Grip Rule Permanent Magnets Biot-Savart Law Magnetic Field of a Current Element Example: Circular Current Loop Right Hand Rule for Current Loop Example: Long Straight Current (*) Permeability Magnetic Fields & Gauss Law Ampere s Law Example: Long Straight Current Example: Infinite Solenoid Summary E and B and Relativity(*) Starred (*) sections will not be included in exam.

3 Moving Charges Contents Previously we have looked at the effect of magnetic fields on moving charges (or currents). The obvious next question is What causes a magnetic field? The previous examples of fields we have seen are symmetric in their causes and effects: A gravitational field is created by a mass, and causes forces on other masses. A charge creates an electric field which causes forces on other charges. This pattern suggests that we should expect that the source of a magnetic field will be a moving charge.

4 Contents And this is easily verified experimentally. Moving a test magnet (e.g. a compass) around a current carrying wire, we observe deflections of the magnet and can thus map the magnetic field lines around the wire. i The result is that the magnetic field lines are found to form closed loops around the wire. The direction of the field lines around the wire is given by the Right Hand Grip rule.

5 Contents Right Hand Grip Rule for Magnetic Field Lines of a Current Using right hand: i place thumb in direction of current wrap fingers around the wire B The magnetic field lines around the wire are in the same direction as fingers. Note: the field lines will generally form squashed loops, rather than the perfect circle drawn here.

6 Permanent Magnets Contents But wait - our fridge magnets don t have a battery, so there is no current to produce a magnetic field! Yes there is! We know atoms consist of charged electrons orbiting the nucleus. The moving electrons form a current. In most atoms, the electron orbits are arranged symmetrically around the nucleus, which means that the total magnetic field produced averages to zero. In the atoms of some elements however (most famously iron, but also nickel, cobalt and a few others), the electron orbits are not symmetrical and the magnetic fields produced do not cancel out. Each atom acts like a tiny bar magnet. If we line the atoms up with their poles in the same direction we can create a larger magnet. Even though it may not always be obvious, every magnetic field is produced by a current.

7 Biot-Savart Law Contents The magnetic field at a point near a current is given by the Biot-Savart Law. i This Law was determined experimentally and gives the magnetic field db due to a small length dl of current i (a dl current element). ˆr r db db = µ 0 idl ˆr 4π r 2 Where ˆr is the unit vector in the direction of the displacement r. The constant µ 0 is called the permeability of the vacuum and is exactly equal to 4π 10 7 N A 2. Of course, finding the field due to all of a current requires integration: B = db, which in general will be difficult to calculate.

8 Biot-Savart: Example 1 Contents Use the Biot-Savart Law to determine the magnetic field strength at the centre of a circular current loop of radius a and current i. As usual, we need to break the loop up into small segments dl as shown, to each of which which we will apply the Biot-Savart law: db = µ 0 idl ˆr 4π r 2 The Right Hand rule gives the direction of the field at the centre to be out of the page, and the magnitude is: a db r i dl db = µ 0 idl 4π a 2

9 Contents The total field strength B at the centre of the loop is B = db = µ 0i 4πa 2 dl = µ 0i 4πa 2 (2πa) = µ 0i 2a Note that in this particular case, the maths turned out to be quite simple due to the symmetry of the current. If we needed to calculate the field at any other point near this loop, the calculation would be much more difficult.

10 Right Hand Rule for Current Loop Contents Right Hand Rule for Magnetic Field Inside Current-Carrying Loop curl fingers in the direction of current thumb indicates direction of magnetic field vector B at centre of loop B i

11 Biot-Savart: Example 2 Contents Use the Biot-Savart Law to determine the magnetic field strength at a point located a perpendicular distance R from an infinitely long straight wire with current i. The method is the same as in the previous example. For one small element of the wire, the field strength db is: db db = µ 0 idx sin θ 4π r 2 i x dx We can use trigonometry to express r and dx in terms of θ and the constant R: R = r sin(180 θ) = r sin θ r = R sin θ R r θ

12 Contents R x = tan(π θ) = tan θ x = R tan θ differentiating gives dx = R tan 2 θ sec2 θdθ = Rdθ sin 2 θ Using these results, the previous expression for db simplifies to: db = µ 0i sin θdθ 4πR And so, the total field strength B is B= π θ=0 db = µ 0i 4πR π 0 sin θdθ = µ 0i 2πR Clearly, using the Biot-Savart Law for a more irregularly shaped current will be challenging!

13 Permeability Contents Given that µ 0 = 4π 10 7, the term µ0 4π equation may seem a little odd. appearing in the Biot-Savart The clue is in the 0, indicating vacuum. If we instead have a material surrounding the current, the Biot-Savart Law is modified to: db = µ idl ˆr 4π r 2 where µ is the permeability of that particular material, which measures how easily it can be magnetized. The permeability of air is very close to the vacuum value, as is concrete, water and other non-magnetic metals such as aluminium and copper. The permeability of iron on the other hand is about 5000µ 0. In this course we will only be considering situations where µ = µ 0.

14 Magnetic Fields & Gauss Law Contents The Biot-Savart Law is applied in a similar fashion to Coulomb s Law for electric fields. It can be used to calculate the field for any arrangement of currents. Currents are always spread out though and so an integration is always required, unlike the simple Coulomb s Law calculations we saw for one or two point charges. With electric fields we were able to use Gauss Law to simplify some symmetrical problems. Will this work with magnetic fields also? Take an imaginary closed surface S surrounding a bar magnet. Because magnetic field lines are always closed loops, then the magnetic flux is φ B = B da = 0 S S N S This will be true for any surface and any field shape.

15 Contents Gauss Law is not useful for calculating magnetic fields. This is because the sum of magnetic charge inside the surface will always be zero. It is not possible to have a single North or South pole (a magnetic monopole) - unlike with electric charges. There is actually no theoretical reason why magnetic monopoles can t exist, and in fact some speculative theories of the universe actually predict that they must exist. Many experiments have been done looking for these monopoles, but to date none have been observed.

16 Ampere s Law Contents All is not lost however! There is a method, which similarly to Gauss Law depends on geometry and symmetry, that we can in some cases use as an easier alternative to Biot-Savart. This is Ampere s Law. We know that a collection of currents will create a magnetic field at all points surounding them. As in the example at right, we now choose an imaginary closed Amperian path P. P I 1 I 2 I 3 I 4 Logically, if we can add up all of the magnetic field directed along this path, then this result should should be proportional to the current inside the path which creates the field. (Recall the very similar reasoning behind Gauss Law)

17 Contents To calculate the sum of fields along the path P, we need to choose a direction (anti-clockwise in the example) and divide the path up into infinitesmal displacement vectors ds. For each of these displacements we calculate the dot product B ds, where B is the magnetic field at that point. The dot product picks out the part of the field in the direction of the path P. Closed Path P ds B I 1 I 2 I 3 I 4 The total field along the path P will be proportional to I P - the total current passing through P: B ds = µ 0 I P P where the permeability of the vacuum is familiar from Biot-Savart: µ 0 = 4π 10 7 N A 2

18 Contents In our example only I 1, I 2 and I 3 contribute to I P. I 4 does not pass through P. (Remember in Gauss Law we similarly ignore charge outside the Gaussian surface) These currents are in differing directions however - how does this affect our sum? Closed Path P ds B I 1 I 2 I 3 I 4 To determine the signs of currents we use our right hand (again!). curl fingers in the direction of the path P thumb will indicate the positive direction for currents In the example, this process indicates that currents out of the page will be positive, so: I P = I 1 + I 3 I 2

19 Contents Applying Amperes Law involves similar difficulties to those seen with Gauss Law. Calculating the integral P B ds is generally impossible, unless we can choose a path P where B is constant and the vectors B and ds are parallel. There are are only a handful of cases where, due to the symmetry of the current, we can do this. Like Gauss Law, Ampere s Law is very important theoretically and is most often used in the derivation of other more practically useful formulas.

20 Example Contents Use Ampere s Law to determine an expression for the magnetic field strength at a point located at a perpendicular distance r from an infintely long straight wire with current i (this is the second of our earlier Biot-Savart examples, so we know the answer!). The key to this calculation is symmetry. A long straight wire will look the same from all perpendicular directions, so we know that the magnetic field lines must form perfect circles around the wire. i i r B(r) end view Again because of the symmetry, we know that the magnitude of the field must depend only on the distance r from the wire.

21 Contents To maintain a constant magnitude B, we need a path P with constant r i.e. a circle. At all points on this path, if we choose an anticlockwise direction, the vectors B and ds are parallel. ds B(r) r i The next calculation should look familiar: B ds = B ds cos 0 (B and ds are parallel everywhere on P) P P = B(r) ds (by symmetry, B is constant on P) P = B(r) (Length of P) = B(r) 2πr (P is a circle)

22 Contents Calculating the current passing through P is easy in this case as there is only one current involved: I P = +i So, applying Ampere s Law: B ds = µ 0 I P P B(r) 2πr = µ 0 i B(r) = µ 0i 2πr Which is the same result that we obtained earlier using Biot-Savart, but with less complicated maths required.

23 Solenoid Contents A solenoid is made up of a single wire wrapped around a cylinder to form a helix. When a current is passed through this wire, the magnetic field produced is exactly the same as that of a bar magnet. i i S N Yet another right hand rule gives the direction of the solenoid field - with fingers in the direction of the current, the thumb indicates the North pole of the solenoid. A magnet that can be switched on and off, solenoids are widely used in electrical machinery (and are often referred to as electromagnets ).

24 Example Contents Use Ampere s Law to determine an expression for the magnetic field strength at a point inside an infinitely long solenoid with current i and n turns of wire per metre. Of course, infinitely long is an approximation, but one that should be valid near the middle of a very long solenoid. B 0 i B i B 0 In this approximation, the magnetic field lines inside the coil will run parallel to the axis of the solenoid, while outside, the field will be very small - approximately zero.

25 Contents We expect the magnitude of the field to depend on the position inside the solenoid, so choose an Amperian path with this constant: B 0 a i b ds i Along the path ab, the field magnitude B is constant and the vectors B and ds are parallel as required. BUT, this path is unsuitable for Ampere s Law because it is not closed!

26 Contents The solution should be familiar. We must build a larger path that is closed and includes ab as a sub-path. This can be done by including sections where B and ds are perpendicular: Closed Path P d ds c ds B 0 ds i a ds b i This new path P is closed, so can be used for Ampere s Law. P For the c b B ds = b a and a d c d B ds + B ds + B ds + b c a d B ds terms we have B ds, so both are zero.

27 Contents In the infinitely long approximation, B = 0 outside the solenoid, so the d term also equals zero. We are left with: c P B ds = = b a b a = B(r) B ds B ds cos 0 (B and ds are parallel on this path) b a ds = B(r) (Length of ab) = B(r) L ab (by symmetry, B is constant on this path)

28 Contents Next we need to determine I P - the total current passing through the path P. The same current i passes through a number of times. How many times? We were told initially that the solenoid has n turns of wire per metre. The path P contains a length L ab of the solenoid. The number of times the wire passes through P must therefore be: d N turns c N = n L ab a L ab b Each turn of the wire carries a current i, so the total current passing through P is: I P = N i = n L ab i

29 Contents Now using Ampere s Law: B ds = µ 0 I P P B(r) L ab = µ o n L ab i B(r) = µ 0 n i Note that, as we should have expected, the length L ab cancels, as it represents the length of an imaginary path. The field B is constant inside the solenoid. Also, the field strength does not depend on the size of the solenoid - only how tightly wound it is.

30 Contents Calculate the magnetic field at a point inside and midway along a solenoid that has 1500 turns of wire in a length of 1.25 m if a current of 2.5 A is flowing. This is not an infinite solenoid, but is long enough that the infinite solenoid formula should be a reasonable approximation when close to the middle of the solenoid. The number of turns per metre, n, of the solenoid is: n = total no. of turns total length = = 1200 Now we can use the formula just found for the infinite solenoid: B = µ 0 n i = 4π = 3.8 mt

31 Other Cases of Ampere s Law Contents There are two other current shapes that can be easily dealt with using Ampere s Law: The toroid which is a single wire wrapped around in a donut shape. An infinite flat current-carrying plate. These cases will be looked at in tutorials. Like the solenoid, there aren t really any variations of these shapes.

32 Summary Contents Magnetic fields are created by moving charges (i.e. electric currents) The basic formula which can always be used to find the field due to a current is the Biot-Savart Law (This is used in the same way that Coulomb s Law is used for electric fields) When there is a high degree of symmetry, we may be able to use Ampere s Law. (Similar to the way that Gauss Law is applied to electric fields)

33 E and B and Relativity (non-examinable) Contents Imagine a student, A, holding two positive charges at rest in each hand. This student will observe a Coulomb force between the charges, explained in terms of an electric field. rest B F v A F + + If student A is inside a moving box, then a second student B will observe the charges to be moving, and will interpret the observed force between them in terms of a magnetic field. Even though they observe different electric and magnetic fields, they do agree on the magnitude of the Lorentz force on the charges: F = q (E + v B)

34 Contents Observers in different inertial reference frames will see different E and B fields but agree on the force produced by them, which is of course the important thing. From this we can conclude that electric and magnetic fields are not separate, but both part of a single electromagnetic field. Electricity and magnetism are very closely linked. Galilean Relativity (i.e. Classical Physics) predicts that the observed forces are different. The mathematics involved in comparing the observations of electromagnetism for different observers was an important inspiration for Einstein s Theory of Special Relativity.