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Fall 2008Physics 231Lecture 8-1 Sources of the Magnetic Field.

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Presentation on theme: "Fall 2008Physics 231Lecture 8-1 Sources of the Magnetic Field."— Presentation transcript:

1 Fall 2008Physics 231Lecture 8-1 Sources of the Magnetic Field

2 Fall 2008Physics 231Lecture 8-2 Magnetic Field of a Point Charge Given a point charge, q, we know that it generates an electric field regardless of whether it is moving or not If the charge is in fact moving then the charge also generates a magnetic field The magnetic field generated by the charge however does not behave in the same way the electric field of the charge does

3 Fall 2008Physics 231Lecture 8-3 Magnetic Field of a Point Charge It is found that the magnetic field is perpendicular to both the velocity of the charge and the unit vector from the charge to the point in question The magnitude of the field is given by where  is the angle between the velocity and unit vector The magnetic field lines form concentric circles about the velocity vector

4 Fall 2008Physics 231Lecture 8-4 Magnetic Field of a Point Charge The direction of the magnetic field is given by one of the right-hand rules Point your thumb, of your right hand, in the direction of the velocity and curl your fingers. The direction of the field is then the direction your fingers curl. The magnetic field for a negative charge will be in the opposite direction. The vector equation for this is

5 Fall 2008Physics 231Lecture 8-5 Magnetic Field of a Current Element If instead of a single charge, we have a current what is the magnetic field then given by We start with the principle of superposition – we add the magnetic fields of the individual charges that make up the current Given n moving charges per unit volume each carrying a charge q, then the total charge in this element is where A is the cross sectional area

6 Fall 2008Physics 231Lecture 8-6 Magnetic Field of a Current Element We then have for dB But from our work on currents we also have that Therefore Or in vector form

7 Fall 2008Physics 231Lecture 8-7 Biot-Savart Law For a complete circuit we then use the integral form of the previous equation This can be a very difficult integral to evaluate We will develop another, much simpler method, shortly The level of difficulty depends upon the variable chosen to integrate over and the presence of any symmetries that can be exploited

8 Fall 2008Physics 231Lecture 8-8 Field of a Straight Wire We set up the problem as shown The integral becomes The result of this integral is In the limit that a >> x

9 Fall 2008Physics 231Lecture 8-9 Magnetic Force Between Two Current Carrying Wires We have two wires carrying currents I and I’ separated by a distance r The first wire carrying current I sets up a magnetic field at the second wire given by

10 Fall 2008Physics 231Lecture 8-10 Magnetic Force Between Two Current Carrying Wires From before we know that a wire carrying a current that is placed in a magnetic field will experience a force per unit length that is given by This is in fact what the second wire will experience And from This force is directed towards the other wire – the force is attractive

11 Fall 2008Physics 231Lecture 8-11 Magnetic Force Between Two Current Carrying Wires If we had started with the wire carrying the current I’, we would have ended up with the same result If the wires were carrying currents in opposite directions, then the force between the wires would be repulsive

12 Fall 2008Physics 231Lecture 8-12 Another circuit for which it is worthwhile to calculate the magnetic field is the circular current loop We start by utilizing the Biot-Savart law witha being the radius of the loop Circular Current Loop

13 Fall 2008Physics 231Lecture 8-13 Circular Current Loop A current element at the top will yield the magnetic field as shown A current element at the bottom will yield a magnetic field of the same magnitude, but directed downward The vertical components will cancel and all that will be left is a horizontal magnetic field of incremental value

14 Fall 2008Physics 231Lecture 8-14 The integral is over just the circumference of the loop as all other variables are constant If there are N loops, turns, then the result is just N times above results Circular Current Loop

15 Fall 2008Physics 231Lecture 8-15 An important point for the field of a current loop is at the center of the loop where x = 0 A current loop consisting of N turns is often referred to as a Solenoid A final caution: These results for a current loop only apply on the axis of the loop Circular Current Loop

16 Fall 2008Physics 231Lecture 8-16 Ampere’s Law Just as we developed an easy way to calculate the electric field in symmetric situations, Gauss’ Law, there is a similar technique we can use for calculating magnetic fields This is known as Ampere’s Law where the integral is around a closed path and I is the total current passing through the area bounded by this path

17 Fall 2008Physics 231Lecture 8-17 There is one big difference to remember between the two methods Gauss’ Law is an integral over a closed surface Whereas Ampere’s Law is an integral over a closed path Ampere’s Law

18 Fall 2008Physics 231Lecture 8-18 Ampere’s Law Given a path around which the integration is to be done, there is a sign convention for the current Determine the direction of the normal, as defined by the sense of movement around the integration path If the current is going in the same direction as the normal, the current is considered to be positive otherwise it is negative

19 Fall 2008Physics 231Lecture 8-19 Long Straight Wire Current in same direction as normal of integration path This is the same as we derived from the Biot-Savart Law

20 Fall 2008Physics 231Lecture 8-20 Current in opposite direction to normal of integration path Again this is the same as we derived from the Biot- Savart Law Long Straight Wire

21 Fall 2008Physics 231Lecture 8-21 Caution Ampere’s Law gives you only the magnetic field due to any currents that cut through the area bounded by the integration path For the indicated integration path, the result of Ampere’s Law is zero

22 Fall 2008Physics 231Lecture 8-22 AB A B CSame 1) For which loop is  B dl the smallest?   Example C Given three currents and their associated loops The currents that are used in Ampere’s Law are the currents that are within the loop that is used for integration Loop A has no current going through it whereas the other two loops do have currents going through them, the integral for Loop A is zero

23 Fall 2008Physics 231Lecture 8-23 B C Same 2) Now compare loops B and C. For which loop is  B dl the greatest?  AB Example C Given three currents and their associated loops The right hand side of Ampere’s law is the same for both loops, therefore the integral for both loops is the same

24 Fall 2008Physics 231Lecture 8-24 The total current going through this loop is the sum of the two current yielding a current of –I. This gives a magnetic that points in the positive x direction at a x I B B B B Example (a) B x (a) < 0 (b) B x (a) = 0 (c) B x (a) > 0 x x x x x x x x 2I2I I a b x y 1) What is the magnetic field B x (a) at point a, just outside the cylinder as shown? A current I flows in an infinite straight wire in the +z direction as shown. A concentric infinite cylinder of radius R carries current 2I in the -z direction. The loop used for application of Ampere’s Law will be outside the cylinder

25 Fall 2008Physics 231Lecture 8-25 This time, the Ampere loop only encloses current I which is in the +z direction — the loop is inside the cylinder! Example (a) B x (b) < 0 (b) B x (b) = 0 (c) B x (b) > 0 2) What is the magnetic field B x (b) at point b, just inside the cylinder as shown? x x x x x x x x 2I2I I a b x y A current I flows in an infinite straight wire in the +z direction as shown. A concentric infinite cylinder of radius R carries current 2I in the -z direction. The current in the cylindrical shell does not contribute to at point b.

26 Fall 2008Physics 231Lecture 8-26 Long Cylindrical Conductor The problem can be divided into two regions Outside the conductor Inside the conductor

27 Fall 2008Physics 231Lecture 8-27 Long Cylindrical Conductor We start with an integration path around the conductor at an arbitrary distance r from the central axis of the conductor Then using Ampere’s Law we then have or Outside the conductor

28 Fall 2008Physics 231Lecture 8-28 Inside the conductor Since we have not been told otherwise, we assume that the current is uniformly distributed across the cross section of the conductor We define a current density We now apply Ampere’s Law around an integration path centered on the cylindrical axis and radius r < R with I being the current going through the area defined by the integration path Long Cylindrical Conductor

29 Fall 2008Physics 231Lecture 8-29 Inside the conductor and So after equating, we then have The magnetic field increases linearly with the distance from the axis of the cylinder Long Cylindrical Conductor

30 Fall 2008Physics 231Lecture 8-30 Inside the wire: (r < a) Outside the wire: ( r > a ) r B a Long Cylindrical Conductor

31 Fall 2008Physics 231Lecture 8-31 Solenoid We choose a path of integration as shown with one part of the path inside the solenoid and one part of the path outside We assume that the path on the outside, cd, is sufficiently far from the solenoid, compared to the diameter of the solenoid, that the field is approximately equal to zero

32 Fall 2008Physics 231Lecture 8-32 We use Ampere’s Law First the Left-Hand-Side a  b: Using the right hand rule for currents, the B field is in the same direction as the path and it is constant, we then have Solenoid

33 Fall 2008Physics 231Lecture 8-33 b  c: Along the part of the path that is inside the solenoid, B is perpendicular to dl, so that contribution is zero, while for the part of the path that is outside of the solenoid, B ~0, so we then have The same is true for d  a c  d: Here B is zero so we again have Solenoid

34 Fall 2008Physics 231Lecture 8-34 Now for the Right-Hand-Side The total current passing through the area defined by the integration path is NI where N is the number of turns and I is the current in the loop, so the right-hand-side is Now equating the left and right-hand-sides we have Or solving for B Solenoid

35 Fall 2008Physics 231Lecture 8-35 Solenoid The magnetic field of a solenoid looks like the following with the exact shape of the field lines being dependent upon the the number of turns, the length, and the diameter of the solenoid


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