 # Physics 1502: Lecture 18 Today’s Agenda Announcements: –Midterm 1 distributed available Homework 05 due FridayHomework 05 due Friday Magnetism.

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Physics 1502: Lecture 18 Today’s Agenda Announcements: –Midterm 1 distributed available Homework 05 due FridayHomework 05 due Friday Magnetism

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! "Brute force"  I "High symmetry"

Magnetic Fields x R r   P I dx Infinite line Circular loop x z R R r dB r z  

Force between two conductors Force on wire 2 due to B at wire 1: Total force between wires 1 and 2: Force on wire 2 due to B at wire 1: Direction: attractive for I 1, I 2 same direction repulsive for I 1, I 2 opposite direction

Lecture 18, ACT 1 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. –What is the magnetic field B z (A) at point A, the midpoint between the two loops? (a) B z (A) < 0 (b) B z (A) = 0 (c) B z (A) > 0

Lecture 18, ACT 1 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. (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?

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. –Why? »Magnitude of B is constant (fct of R only) »Direction of B is parallel to the path. –Current enclosed by path = I –Evaluate line integral in Ampere’s Law:  –Apply Ampere’s Law:

What is the B field at a distance R, with R<a (a: radius of wire)? Choose loop to be circle of radius R, whose edges are inside the wire. –Current enclosed by path = J x Area of Loop B Field inside a Long Wire ? R  I Radius a –Why? »Left Hand Side is same as before.  –Apply Ampere’s Law:

Review: B Field of a Long Wire Inside the wire: (r < a) Outside the wire: (r>a) r B a

Lecture 18, ACT 3 A current I flows in an infinite straight wire in the +z direction as shown. A concentric infinite cylinder of radius R carries current I in the -z direction. –What is the magnetic field B x (a) at point a, just outside the cylinder as shown? 2A (a) B x (a) < 0 (b) B x (a) = 0 (c) B x (a) > 0

Lecture 18, ACT 3 A current I flows in an infinite straight wire in the +z direction as shown. A concentric infinite cylinder of radius R carries current I in the -z direction. 2B (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?

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

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: (n: number of turns per unit length) 

Toroid Toroid defined by N total turns with current i. B=0 outside toroid! (Consider integrating B on circle outside toroid) 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: 

Magnetic Flux Define the flux of the magnetic field through a surface (closed or open) from: Gauss’s Law in Magnetism dS B B

Magnetism in Matter When a substance is placed in an external magnetic field B o, the total magnetic field B is a combination of B o and field due to magnetic moments (Magnetization; M): – B = B o +  o M =  o (H +M) =  o (H +  H) =  o (1+  ) H »where H is magnetic field strength  is magnetic susceptibility Alternatively, total magnetic field B can be expressed as : –B =  m H »where  m is magnetic permeability »  m =  o (1 +  ) All the matter can be classified in terms of their response to applied magnetic field: –Paramagnets  m >  o –Diamagnets  m <  o –Ferromagnets  m >>>  o

Faraday's Law dS B B v B N S v B S N

Induction Effects v v S N v N S N S S N Bar magnet moves through coil  Current induced in coil Change pole that enters  Induced current changes sign Bar magnet stationary inside coil  No current induced in coil Coil moves past fixed bar magnet  Current induced in coil

Faraday's Law Define the flux of the magnetic field through a surface (closed or open) from: Faraday's Law: The emf induced around a closed circuit is determined by the time rate of change of the magnetic flux through that circuit. The minus sign indicates direction of induced current (given by Lenz's Law). dS B B

Faraday’’s law for many loops Circuit consists of N loops: all same area  B magn. flux through one loop loops in “series” emfs add!

Lenz's Law Lenz's Law: The induced current will appear in such a direction that it opposes the change in flux that produced it. Conservation of energy considerations: Claim: Direction of induced current must be so as to oppose the change; otherwise conservation of energy would be violated. »Why??? If current reinforced the change, then the change would get bigger and that would in turn induce a larger current which would increase the change, etc.. v B S N v B N S

Lecture 18, ACT 4 A conducting rectangular loop moves with constant velocity v in the +x direction through a region of constant magnetic field B in the -z direction as shown. – What is the direction of the induced current in the loop? (c) no induced current (a) ccw (b) cw 4A x y

Lecture 18, ACT 4 A conducting rectangular loop moves with constant velocity v in the -y direction away from a wire with a constant current I as shown. What is the direction of the induced current in the loop? 4B (a) ccw (b) cw (c) no induced current x y i

Calculation Suppose we pull with velocity v a coil of resistance R through a region of constant magnetic field B. –What will be the induced current? »What direction? Lenz’ Law  clockwise!! x x x v w x   I –What is the magnitude? »Magnetic Flux: »Faraday’s Law:

 B  E x x x x x r E E E E B Suppose B is increasing into the screen as shown above. An E field is induced in the direction shown. To move a charge q around the circle would require an amount of work = Faraday's law  a changing B induces an emf which can produce a current in a loop. In order for charges to move (i.e., the current) there must be an electric field.  we can state Faraday's law more generally in terms of the E field which is produced by a changing B field. This work can also be calculated from

 B  E Putting these 2 eqns together:  Therefore, Faraday's law can be rewritten in terms of the fields as: x x x x x r E E E E B Note that for E fields generated by charges at rest (electrostatics) since this would correspond to the potential difference between a point and itself. Consequently, there can be no "potential function" corresponding to these induced E fields.

Lecture 18, ACT 5 The magnetic field in a region of space of radius 2R is aligned with the z-direction and changes in time as shown in the plot. –What is sign of the induced emf in a ring of radius R at time t=t 1 ? 5A t B z t 1 X X X X X X X X X X X X X X X X X X X X X X x y X X X X X X X X X X X X X R (a)  < 0 ( E ccw) (b)  = 0 (c)  > 0 ( E cw)

Lecture 18, ACT 5 5B – What is the relation between the magnitudes of the induced electric fields E R at radius R and E 2R at radius 2R ? (a) E 2R = E R (b) E 2R = 2E R (c) E 2R = 4E R t B z t 1 X X X X X X X X X X X X X X X X X X X X X X x y X X X X X X X X X X X X X R

Example An instrument based on induced emf has been used to measure projectile speeds up to 6 km/s. A small magnet is imbedded in the projectile, as shown in Figure below. The projectile passes through two coils separated by a distance d. As the projectile passes through each coil a pulse of emf is induced in the coil. The time interval between pulses can be measured accurately with an oscilloscope, and thus the speed can be determined. (a) Sketch a graph of  V versus t for the arrangement shown. Consider a current that flows counterclockwise as viewed from the starting point of the projectile as positive. On your graph, indicate which pulse is from coil 1 and which is from coil 2. (b) If the pulse separation is 2.40 ms and d = 1.50 m, what is the projectile speed ?

A Loop Moving Through a Magnetic Field  (t) = ?  (t) = ? F(t) = ?

Schematic Diagram of an AC Generator  d dt B      d (cos(  t)) dt   sin(  t))

Schematic Diagram of an DC Generator

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