1 / 34

The Laws of Biot-Savart & Ampere

´. dl. I. Magnetism. The Laws of Biot-Savart & Ampere. Overview of Lecture. Fundamental Law for Calculating Magnetic Field Biot-Savart Law (“brute force”) Ampere’s Law (“high symmetry”) Example: Calculate Magnetic Field of ¥ Straight Wire from Biot-Savart Law from Ampere’s Law

omer
Download Presentation

The Laws of Biot-Savart & Ampere

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. ´ dl I Magnetism The Laws of Biot-Savart & Ampere

  2. Overview of Lecture • Fundamental Law for Calculating Magnetic Field • Biot-Savart Law (“brute force”) • Ampere’s Law (“high symmetry”) • Example: Calculate Magnetic Field of ¥ Straight Wire • from Biot-Savart Law • from Ampere’s Law • Calculate Force on Two Parallel Current-Carrying Conductors Text Reference: Chapter 30.1-4

  3. "Brute force" "High symmetry" Calculation of Electric Field • Two ways to calculate the Electric Field: • Coulomb's Law: • Gauss' Law • What are the analogous equations for the Magnetic Field?

  4. ´ "Brute force" I "High symmetry" Calculation of Magnetic Field • Two ways to calculate the Magnetic Field: • Biot-Savart Law: • Ampere's Law • These are the analogous equations for the Magnetic Field!

  5. dl X r q dB I Biot-Savart Law…bits and pieces (~1819) So, the magnetic field “circulates” around the wire

  6. P r R q x I dx Þ Þ \ Þ Magnetic Field of ¥ Straight Wire • Calculate field at point P using Biot-Savart Law: • Rewrite in terms of R,q: Which way is B?

  7. P r R q x I dx Þ \ Magnetic Field of ¥ Straight Wire

  8. I R (c) B = (m0I)/(2pR) (b) B = (m0I)/(2R) (a) B = 0 Lecture 14, ACT 1 • What is the magnitude of the magnetic field at the center of a loop of radius R, carrying current I?

  9. I r Idx R (c) B = (m0I)/(2pR) (b) B = (m0I)/(2R) (a) B = 0 Lecture 14, ACT 1 • What is the magnitude of the magnetic field at the center of a loop of radius R, carrying current I? • To calculate the magnetic field at the center, we must use the Biot-Savart Law: • Two nice things about calculating B at the center of the loop: • Idxis always perpendicular to r • r is a constant (=R)

  10. dl I R • Evaluate line integral in Ampere’s Law: • Apply Ampere’s Law: Þ Magnetic Field of ¥ Straight Wire • Calculate field at distance R from wire using Ampere's Law: ´ • Choose loop to be circle of radius R centered on the wire in a plane ^ to wire. • Why? • Magnitude of B is constant (fcn of R only) • Direction of B is parallel to the path. • Current enclosed by path =I • Ampere's Law simplifies the calculation thanks to symmetry of the current! ( axial/cylindrical )

  11. y a x x x b 2I x x I x x x (c) Bx(a) > 0 (b) Bx(a) = 0 (a) Bx(a) < 0 x (c) Bx(b) > 0 (b) Bx(b) = 0 (a) Bx(b) < 0 Lecture 14, ACT 2 • A currentI 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. • What is the magnetic field Bx(a) at point a, just outside the cylinder as shown? • What is the magnetic field Bx(b) at point b, just inside the cylinder as shown?

  12. y a x x x b 2I x x I x x x (c) Bx(a) > 0 (b) Bx(a) = 0 (a) Bx(a) < 0 x B x B I B B Lecture 14, ACT 2 • A currentI 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. • What is the magnetic field Bx(a) at point a, just outside the cylinder as shown? • This situation has massive cylindrical symmetry! • Applying Ampere’s Law, we see that the field at point a must just be the field from an infinite wire with current I flowing in the -z direction!

  13. y a x x x b 2I x x I x x x (c) Bx(a) > 0 (b) Bx(a) = 0 (a) Bx(a) < 0 x (c) Bx(b) > 0 (b) Bx(b) = 0 (a) Bx(b) < 0 • A currentI 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. • What is the magnetic field Bx(a) at point a, just outside the cylinder as shown? Lecture 14, ACT 2 What is the magnetic field Bx(b) at point b, just inside the cylinder as shown? • Just inside the cylinder, the total current enclosed by the Ampere loop will be I in the +z direction! • Therefore, the magnetic field at b will just be minus the magnetic field at a!!

  14. F ´ Ib d L Ia Þ Force on b = Ib d L Ia ´ Þ Force on a = F Force on 2 ParallelCurrent-Carrying Conductors • Calculate force on length L of wire b due to field of wire a: • The field at b due to a is given by: • Calculate force on length L of wire a due to field of wire b: The field at a due to b is given by:

  15. I y I (a) Fx < 0 (c) Fx > 0 (b) Fx = 0 x Lecture 14, ACT 3 • A current I flows in the positive y direction in an infinite wire; a current I also flows in the loop as shown in the diagram. • What is Fx, net force on the loop in the x-direction?

  16. I Ftop y Fright X Fleft I Fbottom (a) Fx < 0 (c) Fx > 0 (b) Fx = 0 x • The direction of the magnetic field at the current loop is in the -z direction. • The forces on the top and bottom segments of the loop DO indeed cancel!! • A current I flows in the positive y direction in an infinite wire; a current I also flows in the loop as shown in the diagram. • What is Fx, net force on the loop in the x-direction? Lecture 14, ACT 3 • You may have remembered from a previous ACT that the net force on a current loop in a constant magnetic field is zero. • However, the magnetic field produced by the infinite wire is not a constant field!! • The forces on the left and right segments of the loop DO NOT cancel!! • The left segment of the loop is in a larger magnetic field. • Therefore, Fleft > Fright

  17. x • x dB r q x R x q x z x z x R r x dB x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x a r Examples of Magnetic Field Calculations

  18. Overview of Lecture • Calculate Magnetic Fields • Inside a Long Straight Wire • Infinite Current Sheet • Solenoid • Toroid • Circular Loop Text Reference: Chapter 30.1-5

  19. Integral around a path … hopefully a simple one Current “enclosed” by that path ´ I Today is Ampere’s Law Day "High symmetry"

  20. x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x a r Þ Þ B Field Insidea Long Wire • Suppose a total current I flows through the wire of radius a into the screen as shown. • Calculate B field as a fcn of r, the distance from the center of the wire. • Bfield is only a fcn of rÞ take path to be circle of radius r: • Current passing through circle: • Ampere's Law:

  21. a B r B Field of aLong Wire • Inside the wire: (r < a) • Outside the wire: (r>a)

  22. I I a 3a 2a 3a 1A (a) BL(6a)< BR(6a) (b) BL(6a)= BR(6a) (c) BL(6a)> BR(6a) • What is the relation between the magnetic field at R = 2a for the two cases (L=left, R=right)? 1B (a) BL(2a)< BR(2a) (b) BL(2a)= BR(2a) (c) BL(2a)> BR(2a) Lecture 15, ACT 1 • Two cylindrical conductors each carry current I into the screen as shown.The conductor on the left is solid and has radius R=3a. The conductor on the right has a hole in the middle and carries current only between R=a and R=3a. • What is the relation between the magnetic field at R = 6a for the two cases (L=left, R=right)?

  23. I I a 3a 2a 3a 1A (a) BL(6a)< BR(6a) (b) BL(6a)= BR(6a) (c) BL(6a)> BR(6a) Lecture 15, ACT 1 • Two cylindrical conductors each carry current I into the screen as shown.The conductor on the left is solid and has radius R=3a. The conductor on the right has a hole in the middle and carries current only between R=a and R=3a. • What is the relation between the magnetic field at R = 6a for the two cases (L=left, R=right)? • Ampere’s Law can be used to find the field in both cases. • The Amperian loop in each case is a circle of radius R=6a in the plane of the screen. • The field in each case has cylindrical symmetry, being everywhere tangent to the circle. • Therefore the field at R=6a depends only on the total current enclosed!! • In each case, a total current I is enclosed.

  24. I I a 3a 2a 3a 1A (a) BL(6a)< BR(6a) (b) BL(6a)= BR(6a) (c) BL(6a)> BR(6a) • What is the relation between the magnetic field at R = 2a for the two cases (L=left, R=right)? 1B (a) BL(2a)< BR(2a) (b) BL(2a)= BR(2a) (c) BL(2a)> BR(2a) Lecture 15, ACT 1 • Two cylindrical conductors each carry current I into the screen as shown.The conductor on the left is solid and has radius R=3a. The conductor on the right has a hole in the middle and carries current only between R=a and R=3a. • What is the relation between the magnetic field at R = 6a for the two cases (L=left, R=right)? • Once again, the field depends only on how much current is enclosed. • For the LEFT conductor: • For the RIGHT conductor:

  25. y x x x x x x x x x x x x x constant constant • • \ Þ B Field of ¥ Current Sheet • Consider an ¥ sheet of current described by n wires/length each carrying current i into the screen as shown. Calculate the B field. • What is the direction of the field? • Symmetry Þ y direction! • Calculate using Ampere's law using a square of side w:

  26. L • A solenoid is defined by a current I flowing through a wire which is wrapped n turns per unit length on a cylinder of radius a and length L. a B Field of a Solenoid • A constant magnetic field can (in principle) be produced by an ¥ sheet of current. In practice, however, a constant magnetic field is often produced by a solenoid. • Ifa <<L, the B field is to first order contained within the solenoid, in the axial direction, and of constant magnitude. In this limit, we can calculate the field using Ampere's Law.

  27. x x x x x • • • • • x x x x x • • • • • Þ B Field of a ¥ Solenoid • To calculate the B field of the ¥ solenoid using Ampere's Law, we need to justify the claim that the B field is 0 outside the solenoid. • To do this, view the ¥ solenoid from the side as 2 ¥ current sheets. • The fields are in the same direction in the region between the sheets (inside the solenoid) and cancel outside the sheets (outside the solenoid). • Draw square path of side w:

  28. • • • • • x x x • • x x x x x • x • r x x x x • • x x x • • • • B • Þ Toroid • Toroid defined by N total turns with current i. • To find B inside, consider circle of radiusr, centered at the center of the toroid. • B=0 outside toroid! (Consider integrating B on circle outside toroid) Apply Ampere’s Law:

  29. • dB r q R q z z R r dB x Circular Loop • Circular loop of radius R carries current i. Calculate B along the axis of the loop: • Magnitude of dB from element dl: • What is the direction of the field? • Symmetry Þ B in z-direction. Þ

  30. • dB r q R q z z R r dB x Circular Loop • Note the form the field takes for z>>R: Þ • Expressed in terms of the magnetic moment: note the typical dipole field behavior! Þ

  31. 2A (c) Bz(A) > 0 (b) Bz(A) = 0 (a) Bz(A) < 0 • What is the magnetic field Bz(B) at point B, just to the right of the right loop? 2B (c) Bz(B) > 0 (b) Bz(B) = 0 (a) Bz(B) < 0 Lecture 15, ACT 2 • Equal currentsI flow in identical circular loops as shown in the diagram. The loop on the right(left) carries current in the ccw(cw) direction as seen looking along the +z direction. • What is the magnetic field Bz(A) at point A, the midpoint between the two loops?

  32. 2A Lecture 15, ACT 2 • Equal currentsI flow in identical circular loops as shown in the diagram. The loop on the right(left) carries current in the ccw(cw) direction as seen looking along the +z direction. • What is the magnetic field Bz(A) at point A, the midpoint between the two loops? (c) Bz(A) > 0 (b) Bz(A) = 0 (a) Bz(A) < 0 • The right current loop gives rise to Bz <0 at point A. • The left current loop gives rise to Bz >0 at point A. • From symmetry, the magnitudes of the fields must be equal. • Therefore, B(A) = 0!

  33. 2A 2B Lecture 15, ACT 2 • Equal currentsI flow in identical circular loops as shown in the diagram. The loop on the right(left) carries current in the ccw(cw) direction as seen looking along the +z direction. • What is the magnetic field Bz(A) at point A, the midpoint between the two loops? (c) Bz(A) > 0 (b) Bz(A) = 0 (a) Bz(A) < 0 • What is the magnetic field Bz(B) at point B, just to the right of the right loop? (c) Bz(B) > 0 (b) Bz(B) = 0 (a) Bz(B) < 0 • The signs of the fields from each loop are the same at B as they are at A! • However, point B is closer to the right loop, so its field wins!

  34. 1 » z 3 Circular Loop R B z 0 0 z

More Related