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Examples of Magnetic Field r Calculations Bgd — l 0 I dl I xxxxx xxxxxxxx xxxxxxxxx r xxxxxxxx 1 a xxxxx B Field by Ampere’s Law "High symmetry" r Bgd — l 0 I Integral around a path … Net Current ―enclosed‖ by hopefully a simple one that path I 2 B-field of Straight Wire, reprise • Calculate field at distance R from wire using Ampere's Law: dl Choose loop to be circle of radius R centered on the I R wire in a plane ^ to wire. Why? Magnitude of B is constant (function of R only) Direction of B is parallel to the path. r Evaluate line integral in Ampere’s Law: Bg — dl B(2š R) Current enclosed by path = I Apply Ampere’s Law: 0 I B 2 RB 0 I 2š R Ampere's Law simplifies the calculation thanks to 3 symmetry of the current! (axial/cylindrical) Question Two identical loops are placed in proximity to two identical current carrying wires. A B r Bgd For which loop is — l the greatest? A B Same 4 Question Now compare loops B and C. r For which loop is Bgd — l C the greatest? B B C Same 5 Question y A current I flows in an infinite straight a wire in the +z direction as shown. A x x b x concentric infinite cylinder of radius R carries current 2I in the -z direction. x x 2I – What is the magnetic field Bx(a) at x I x point a, just outside the cylinder as x shown? x (a) Bx(a) < 0 (b) Bx(a) = 0 (c) Bx(a) > 0 • This situation has cylindrical symmetry • By 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! B B x I B 6 B Question y A current I flows in an infinite straight wire a in the +z direction as shown. A concentric x x b x infinite cylinder of radius R carries current 2I in the -z direction. x x 2I – What is the magnetic field Bx(b) at point x I x b, just inside the cylinder as shown? x x (a) Bx(b) < 0 (b) Bx(b) = 0 (c) Bx(b) > 0 • This time, the Ampere loop only encloses current I in the +z direction — the loop is inside the cylinder! • The current in the cylindrical shell does not contribute to at point b. 7 B Field Inside a Long Wire xxxxx • Suppose a total current I flows through the wire of radius a into the screen as xxxxxxxx shown. xxxxxxxxx • Calculate B field as a function of r, the r xxxxxxxx distance from the center of the wire. a xxxxx • B field is only a function of r take path r to be circle of radius r: Bgd — l B(2 š r) r2 • Current passing through circle: I enclosed 2I a r r 0 I r • Ampere's Law: Bgd — l o Ienclosed B 2 š a2 8 B Field of a Long Wire • Inside the wire: (r < a) 0 I r a a B 2 a2 • Outside the wire: B B (r>a) 0 I B 2 r r r 9 Question • Two cylindrical conductors each I I 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 a hole in the middle and carries 3a 3a 2a 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)? (a) BL(6a)< BR(6a) (b) BL(6a)= BR(6a) (c) BL(6a)> BR(6a) • Use Ampere’s Law in both cases by drawing a loop in the plane of the screen at R=6a • Both fields have cylindrical symmetry, so they are tangent to the loop at all points, thus the field at R=6a only depends on current enclosed 10 • Ienclosed = I in both cases Question • Two cylindrical conductors each carry I I 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 a carries current only between R=a and 3a 3a 2a R=3a. – What is the relation between the magnetic field at R = 2a for the two cases (L=left, R=right)? (a) BL(2a)< BR(2a) (b) BL(2a)= BR(2a) (c) BL(2a)> BR(2a) Again, field only depends upon current enclosed... LEFT cylinder: RIGHT cylinder: (2a)2 4 (2a)2 a 2 3 IL I I IR I I (3a)2 9 (3a)2 a 2 8 11 B Field of Current Sheet • Consider an sheet of current described by x n wires/length each carrying current i into the x screen as shown. Calculate the B field. x w x x • What is the direction of the field? x x • Symmetry vertical direction x x • Calculate using Ampere's law for a square of x x side w: x r constant Bgd constant • — l Bw 0 Bw 0 2Bw • I nwi r 0 ni Bgd therefore, — l 0 I B 2 12 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. • A solenoid is defined by a current i flowing L through a wire that is wrapped n turns per unit length on a cylinder of radius a and length L. a To correctly calculate the B-field, we should use Biot-Savart, and add up the field from the different loops. • 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. 13 B Field of an Solenoid • To calculate the B field of the solenoid using Ampere's Law, we need to justify the claim that the B field is nearly 0 outside the solenoid (for an solenoid the B field is exactly 0 outside). • To do this, view the solenoid from the xxxxxxxxxxx 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: xxxxxxxx r • •••••••••• Bgd — l Bw B 0 ni Note: B Amp I nwi Length 14 Question A current carrying wire is wrapped around an iron core, forming an electro-magnet. Which direction does the magnetic field point inside the iron core? a) left b) right c) up d) down e) out of the screen f) into the screen Which side of the solenoid should be labeled as the magnetic north pole? a) right b) left 15 Right Hand Rule Use the right hand rule to find the B-field: wrap your fingers in the direction of the current, the B field points in the direction of the thumb (to the left). Since the field lines leave the left end of solenoid, the left end is the north pole. 16 Solenoids The magnetic field of a solenoid is essentially identical to that of a bar magnet. The big difference is that we can turn the solenoid on and off ! It attracts/repels other permanent magnets; it attracts ferromagnets, etc. 17 Toroid • • • • • • Toroid defined by N total turns • xx x • x x with current i. x x • B=0 outside toroid! (Consider • x r x • x x integrating B on circle outside toroid) • xx x • x x • • • To find B inside, consider circle of radius r, centered at the center of the toroid. • • B• r Bgd — l B(2 š r) I Ni r 0 Ni Apply Ampere’s Law: Bgd — l 0 I B 2š r 18 Magnetic Materials Material is made up of atoms. 2 r For an electron v e T q e ev L I t T 2 r r ev e e r r IA 2 r r 2 2m rmv L 2m r r r where L is the angular momentum r p So a material object is full of magnetic moments --- -- which generally cancel due to no preferred direction. 19 Magnetization & H When a magnetic field is applied the symmetry is broken And there is a direction along which the line up For our toroid let B0 be the magnetic field when the windings are empty and B is the field when the space is filled with material. B B0 Bm M is the magnetization or r total magnetic moment per unit volume Bm 0 0 M i V r r Define another vector H such that B 0 H M H is due to the current and does not depend on the presence of material. 20 Toroid • • • • • • Toroid defined by N total turns with • xx x • x x current i. x x • B=0 outside toroid! (Consider integrating • x r x • B on circle outside toroid) x x • xx x x • x • • • To find B inside, consider circle of radius r, centered at the center of the toroid. • • B• r Bgd — l B(2 š r) I Ni 2 rni 0 Ni B 2š r 0 ni r Bgd Apply Ampere’s Law: — l I 21 Materials r Recall there are three types of materials M H •Diamagnetic 0 and small : 10 5 •Paramagnetic 0 and small : 10 5 •Ferromagnetic 0 very large : 5000 22

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