# Physics 1304: Lecture 12, Pg 1 The Laws of Biot-Savart & Ampere  dl I.

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Physics 1304: Lecture 12, Pg 1 The Laws of Biot-Savart & Ampere  dl I

Physics 1304: Lecture 12, Pg 2 Overview of Lecture l 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 l Calculate Force on Two Parallel Current-Carrying Conductors Text Reference: Chapter 30.1-4

Physics 1304: Lecture 12, Pg 3 Calculation of Electric Field "Brute force" "High symmetry" l Two ways to calculate the Electric Field: Coulomb's Law: Gauss' Law l What are the analogous equations for the Magnetic Field?

Physics 1304: Lecture 12, Pg 4 Calculation of Magnetic Field "High symmetry" "Brute force"  I l Two ways to calculate the Magnetic Field: Biot-Savart Law: Ampere's Law l These are the analogous equations for the Magnetic Field!

Physics 1304: Lecture 12, Pg 5 Biot-Savart Law… bits and pieces I dl dB X r  So, the magnetic field “circulates” around the wire

Physics 1304: Lecture 12, Pg 6 Magnetic Field of  Straight Wire Calculate field at point P using Biot-Savart Law: Rewrite in terms of R,  : x R r  P I dx     Which way is B?

Physics 1304: Lecture 12, Pg 7 Magnetic Field of  Straight Wire   x R r  P I dx

Physics 1304: Lecture 12, Pg 8 Lecture 14, ACT 1 What is the magnitude of the magnetic field at the center of a loop of radius R, carrying current I ? (a) B = 0 (b) B    R  (c) B    R  R I

Physics 1304: Lecture 12, Pg 9 Lecture 14, ACT 1 What is the magnitude of the magnetic field at the center of a loop of radius R, carrying current I ? (a) B = 0 (b) B    R  (c) B    R  To calculate the magnetic field at the center, we must use the Biot- Savart Law: Idx r Two nice things about calculating B at the center of the loop: Idx is always perpendicular to r r is a constant (=R) R I

Physics 1304: Lecture 12, Pg 10 Magnetic Field of  Straight Wire Calculate field at distance R from wire using Ampere's Law: Ampere's Law simplifies the calculation thanks to symmetry of the current! ( axial/cylindrical )  dl  R I Choose loop to be circle of radius R centered on the wire in a plane  to wire. çWhy? »Magnitude of B is constant (fcn of R only) »Direction of B is parallel to the path. çEvaluate line integral in Ampere’s Law:  Current enclosed by path = I çApply Ampere’s Law:

Physics 1304: Lecture 12, Pg 11 Lecture 14, ACT 2 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. çWhat is the magnetic field B x (a) at point a, just outside the cylinder as shown? (a) B x (b) < 0 (b) B x (b) = 0 (c) B x (b) > 0 – What is the magnetic field B x (b) at point b, just inside the cylinder as shown? (a) B x (a) < 0 (b) B x (a) = 0 (c) B x (a) > 0 x x x x x x x x 2I I a b x y

Physics 1304: Lecture 12, Pg 12 Lecture 14, ACT 2 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. l What is the magnetic field B x (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! x I B B B B x x x x x x x x 2I I a b x y (a) B x (a) < 0 (b) B x (a) = 0 (c) B x (a) > 0

Physics 1304: Lecture 12, Pg 13 Lecture 14, ACT 2 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!! x x x x x x x x 2I 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. l What is the magnetic field B x (a) at point a, just outside the cylinder as shown? (a) B x (a) < 0 (b) B x (a) = 0 (c) B x (a) > 0 (a) B x (b) < 0 (b) B x (b) = 0 (c) B x (b) > 0 What is the magnetic field B x (b) at point b, just inside the cylinder as shown?

Physics 1304: Lecture 12, Pg 14 Question How do we check this result?? i.e. expect B from wire to be proportional to I /R. l Measure FORCE on current-carrying wire due to the magnetic field PRODUCED by ANOTHER current carrying wire! How does force depend on currents and separation? F d IaIa IbIb

Physics 1304: Lecture 12, Pg 15 Force on 2 Parallel Current-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: l Calculate force on length L of wire a due to field of wire b: The field at a due to b is given by:  F  F L d IbIb IaIa L d IbIb IaIa  Force on b =  Force on a =

Physics 1304: Lecture 12, Pg 16 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. ç What is F x, net force on the loop in the x- direction? (a) F x < 0 (b) F x = 0 (c) F x > 0 I I x y

Physics 1304: Lecture 12, Pg 17 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!! F left F right X The direction of the magnetic field at the current loop is in the -z direction. F top F bottom The forces on the top and bottom segments of the loop DO indeed cancel!! 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, F left > F right I I x y 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. ç What is F x, net force on the loop in the x- direction? (a) F x < 0 (b) F x = 0 (c) F x > 0

Physics 1304: Lecture 12, Pg 18 Examples of Magnetic Field Calculations 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 ra x x x x x x x x x x x x x z R R r r dB z  

Physics 1304: Lecture 12, Pg 19 Overview of Lecture l Calculate Magnetic Fields çInside a Long Straight Wire çInfinite Current Sheet çSolenoid çToroid çCircular Loop Text Reference: Chapter 30.1-5

Physics 1304: Lecture 12, Pg 20 Today is Ampere’s Law Day "High symmetry"  I Integral around a path … hopefully a simple one Current “enclosed” by that path

Physics 1304: Lecture 12, Pg 21 B Field Inside a Long Wire 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 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. B field is only a fcn of r  take path to be circle of radius r: Current passing through circle: r Ampere's Law:  

Physics 1304: Lecture 12, Pg 22 B Field of a Long Wire Inside the wire: (r < a) Outside the wire: (r>a) r B a

Physics 1304: Lecture 12, Pg 23 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. ç What is the relation between the magnetic field at R = 6a for the two cases (L=left, R=right)? (a) B L (6a)< B R (6a) (b) B L (6a)= B R (6a) (c) B L (6a)> B R (6a) (a) B L (2a)< B R (2a) (b) B L (2a)= B R (2a) (c) B L (2a)> B R (2a) What is the relation between the magnetic field at R = 2a for the two cases (L=left, R=right)? 1B 1A 3a a I I 2a

Physics 1304: Lecture 12, Pg 24 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. ç What is the relation between the magnetic field at R = 6a for the two cases (L=left, R=right)? (a) B L (6a)< B R (6a) (b) B L (6a)= B R (6a) (c) B L (6a)> B R (6a) 1A 3a a I I 2a 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.

Physics 1304: Lecture 12, Pg 25 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. ç What is the relation between the magnetic field at R = 6a for the two cases (L=left, R=right)? (a) B L (6a)< B R (6a) (b) B L (6a)= B R (6a) (c) B L (6a)> B R (6a) (a) B L (2a)< B R (2a) (b) B L (2a)= B R (2a) (c) B L (2a)> B R (2a) What is the relation between the magnetic field at R = 2a for the two cases (L=left, R=right)? 1B 1A 3a a I I 2a For the LEFT conductor: Once again, the field depends only on how much current is enclosed. For the RIGHT conductor:

Physics 1304: Lecture 12, Pg 26 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. x x x x x x x x x x x x x y What is the direction of the field? Symmetry  y direction! Calculate using Ampere's law using a square of side w:   constant

Physics 1304: Lecture 12, Pg 27 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. If a << 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. 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

Physics 1304: Lecture 12, Pg 28 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. x x xxx The fields are in the same direction in the region between the sheets (inside the solenoid) and cancel outside the sheets (outside the solenoid). x x xxx Draw square path of side w: 

Physics 1304: Lecture 12, Pg 29 Toroid Toroid defined by N total turns with current i. B=0 outside toroid! (Consider integrating B on circle outside toroid) l To find B inside, consider circle of radius r, centered at the center of the toroid. x x x x x x x x x x x x x x x x r B  Apply Ampere’s Law:

Physics 1304: Lecture 12, Pg 30 Circular Loop x z R R Circular loop of radius R carries current i. Calculate B along the axis of the loop: Magnitude of dB from element dl: r dB r z   What is the direction of the field? Symmetry  B in z-direction. 

Physics 1304: Lecture 12, Pg 31 Circular Loop  Note the form the field takes for z>>R: Expressed in terms of the magnetic moment:  note the typical dipole field behavior! x z R R r r dB z  

Physics 1304: Lecture 12, Pg 32 Lecture 15, ACT 2 Equal currents I 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. çWhat is the magnetic field B z (A) at point A, the midpoint between the two loops? (a) B z (B) < 0 (b) B z (B) = 0 (c) B z (B) > 0 2B – What is the magnetic field B z (B) at point B, just to the right of the right loop? (a) B z (A) < 0 (b) B z (A) = 0 (c) B z (A) > 0 2A

Physics 1304: Lecture 12, Pg 33 Lecture 15, ACT 2 Equal currents I 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. çWhat is the magnetic field B z (A) at point A, the midpoint between the two loops? 2A (a) B z (A) < 0 (b) B z (A) = 0 (c) B z (A) > 0 The right current loop gives rise to B z <0 at point A. The left current loop gives rise to B z >0 at point A. From symmetry, the magnitudes of the fields must be equal. Therefore, B(A) = 0!

Physics 1304: Lecture 12, Pg 34 Lecture 15, ACT 2 Equal currents I 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. çWhat is the magnetic field B z (A) at point A, the midpoint between the two loops? 2A2B (a) B z (B) < 0 (b) B z (B) = 0 (c) B z (B) > 0 – What is the magnetic field B z (B) at point B, just to the right of the right loop? (a) B z (A) < 0 (b) B z (A) = 0 (c) B z (A) > 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!

Physics 1304: Lecture 12, Pg 35 Circular Loop R B z z 0 0  1 z 3

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