Physics 1502: Lecture 18Today’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

Rr

P

Idx

• Infinite line

• Circular loop

x

•

z

R

R r dB

r

z

dB

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 I1, I2 same direction

repulsive for I1, I2 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 Bz(A)

at point A, the midpoint between the two loops?

(a) Bz(A) < 0 (b) Bz(A) = 0 (c) Bz(A) > 0

x

o x

o

z

I I

A B

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) Bz(B) < 0 (b) Bz(B) = 0 (c) Bz(B) > 0

– What is the magnetic field Bz(B) at point B, just to the right of the right loop?

x

o x

o

z

I I

A B

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 RI

• 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 ?

RI

Radius a

– Why?» Left Hand Side is same as before.

– Apply Ampere’s Law:

Review: B Field of aLong Wire

B = μ0 I

2 π ra2

• Inside the wire: (r < a)

• Outside the wire: (r>a)

B = μ0 I

2 πr

0 4

x =

y =

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 Bx(a) at

point a, just outside the cylinder as shown?

2A

(a) Bx(a) < 0 (b) Bx(a) = 0 (c) Bx(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) Bx(b) < 0 (b) Bx(b) = 0 (c) Bx(b) > 0

– What is the magnetic field Bx(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.

xxx xx

•• • ••• The fields are in the same direction in the

region between the sheets (inside the solenoid) and cancel outside the sheets (outside the solenoid).

xxx xx

•• • ••• Draw square path of side w:

(n: number ofturns per unitlength)

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 FluxDefine 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 Bo,

the total magnetic field B is a combination of Bo and field due to magnetic moments (Magnetization; M):

– B = Bo + μoM = μo (H +M) = μo (H + H) = μo (1+) H

» where H is magnetic field strength is magnetic susceptibility

• Alternatively, total magnetic field B can be expressed as:– B = μm H

» where μm is magnetic permeability» μm = μo (1 + )

• All the matter can be classified in terms of their response to applied magnetic field:

– Paramagnets μm > μo

– Diamagnets μm < μo

– Ferromagnets μm >>> μo

Faraday's Law

dS

B Bv

BN S

v

BS N

Induction Effects

v

vS N

vN 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

BS N

v

BN 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 X X X X X X X X X X XX X X X X X X X X X X XX X X X X X X X X X X XX X X X X X X X X X X X

v

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

v

I

x

yi

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 x x x

x x x x x x

x x x x x x

x x x x x x

vw

x

I

– What is the magnitude?

» Magnetic Flux:

» Faraday’s Law:

B Ex 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 x x x x 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 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 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=t1?

5A

t

Bz

t1

X X X X X X X X X X XX X X X X X X X

X X X X X X X X Xx

y

X X X X X X X X XX X X X X X 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 ER at radius R and E2R at radius 2R ?

(a) E2R = ER (b) E2R = 2ER (c) E2R = 4ER

t

Bz

t1

X X X X X X X X X X XX X X X X X X X

X X X X X X X X Xx

y

X X X X X X X X XX X X X X X X X X X X X X X X

X X X X

R

ExampleAn 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

dtB

d (cos( t))dt

sin( t))

Schematic Diagram of an DC Generator