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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

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The Laws of Biot-Savart & Ampere

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  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

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