Chapter 21: Magnetic Forces and Fields Notes/Cha… · 21.5/2: B = 0.3 T, v = 45 m/s, q = 8.4 μC....

Chapter 21: Magnetic Forces and Fields

• Magnetic fields

• The force on a moving charge and on a current in a magnetic field

• The torque on a coil in a magnetic field. Electric motors

• Magnetic field produced by a current. Ampere’s law

• Omit 21.9, magnetic materials

1Wednesday, February 6, 2008

Magnetic poles, north and south

Magnetic poles occur only as N-S pairs, unlike electrical charges that exist separately as + and – entities.

Existence of a “magnetic monopole”?

Only a theoretical possibility at present.


The north pole (“north-seeking pole”) of a compass needle points along the magnetic field lines toward the south pole of a magnet

Magnetic field lines form a pattern similar to an electric dipole.

Compass needle points along magnetic field line toward the magnetic south pole


Magnetic field lines of the earth

⇒ the north pole of the earth is a magnetic south pole...


Force on a positive charge moving in a magnetic field

Magnetic field

thuMb (Motion of + charge)

Palm (Push, i.e. force)

Fingers (Field)

Right hand rule F in Newtons

q in Coulombs

v in m/s

B in Tesla (T)

Earth’s magnetic field ! 10-4 T

Force is at right angles to both!v and !B

F = qvBsin!!B

!v

!q


21.C5: Determine whether each particle is positively or negatively charged, or neutral.

!F

!F

#1: positive

#2: neutral

#3: negative


21.5/2: B = 0.3 T, v = 45 m/s, q = 8.4 μC. Find the force on the charge in the three cases below.

F = qvBsin!, qvB= 1.13!10"4 N

a) != 30!, F = 0.57"10#4 N, into screen

b) != 90!, F = 1.13"10#4 N, into screen

c) != 150!, F = 0.57"10#4 N, into screen7Wednesday, February 6, 2008

Motion of a charge in a magnetic field

The force is always perpendicular to

the velocity (and displacement), so the force does no work and the charge moves at constant speed. The charge follows a circular path if the magnetic field is uniform.

Motion of a charge in an electric field

There is in general a component of the force in the direction of the displacement, so the force does work and the speed of the charge changes.


Motion of a charge in a uniform magnetic field

As B is at right angles to v:

F = q v B

F always at right angles to velocity⇒ it does no work, v is constant

The circular path has radius determined by the centripetal force provided by the magnetic field –

Fc =mv2

r= qvB

So, r =mv

qB


Prob. 21.11/12:

a) Is q positive or negative?

Right hand rule: q is negative

b) If the magnitude of q is 8.2"10-4 C, what is the mass of the particle?

r =mv

qB

So, m =qBr

v=8.2!10"4!0.48!960

140

= 2.7!10"3 kg

v = 140 m/s

B = 0.48 T

r = 960 m


Mass spectrometer, to measure mass of ions

Removes an electron from neutral atoms, forming singly-charged positive ions

m, q = e

!F

v ! 0

and1

2mv

2 = eV

so, v=!2eV/m

m=qBr

v=eBr

v

m=eB

2r2

2V

In the magnetic field:

KE = eV

–V


Mass Spectrum

m=eB

2r2

2V

Isotopes of neon

⇐ atomic mass of isotopes of neon


Prob. 21.18: The ion source in a mass spectrometer produces both singly and doubly ionized species, X+ and X2+. The difference in mass is too small to detect.

Both species are accelerated through the same electric potential difference and experience the same magnetic field, which causes them to move on circular paths.

Find the ratio of the radii of the paths.

For a particle of charge q : m =qB2r2

2V


Velocity SelectorMagnetic Field

The charge +q feels an upward force due to the magnetic field.

Electric Field

The charge +q feels a downward force due to the electric field.

With both the B and E fields present, the net force on the charge is zero if:

qvB = qE, that is,

F = qE

F = qvB

v = E/B ⇐ measurement of velocity, independent of q, m


Force on a current-carrying conductor

Current is a flow of charge.

Imagine + charges moving along the conductor in the direction of the current, I.

There is a force F acting on the charges moving along the conductor that is due to the magnetic field.

The direction of the force is given by the right hand rule.



A charge Δq flows at speed v along a length

L of wire in time Δt.

I =!q

!t

The force on the charge Δq is:

F = !q v Bsin"=!!q

!t

"(v !t) Bsin"

F = ILBsin!

!v

and L = vΔt

L

Δq

I


By how much does the weight of the coil of wire seem to increase when the current is turned on in the coil?

B = 0.2 T, I = 8.5 A

I

F


Force per turn on lower end of loop:

F = I L B, downwards

There are N = 125 turns in the coil, so total force is:

F = N I L B = 125 I L B

B = 0.2 T, I = 8.5 A

125 turns

F = 125 " (8.5 A) " (0.015 m) " (0.2 T) = 3.19 N

I

F


01/02/08 11:58 AMMasteringPhysics

Page 1 of 2http://session.masteringphysics.com/myct

Power Dissipation in Resistive Circuit Conceptual QuestionA single resistor is wired to a battery as shown in the diagram below. Define the total power dissipated

by this circuit as .

Jim BirchallNow, a second identical resistor is wired in series with the first resistor as shown in the second diagram tothe left .

Part AWhat is the power, in terms of , dissipated by this circuit?

Express your answer in terms of .

Mastering Physics printing bug!

Problem 1, assignment 2:

If you print out the problem with the usual print command, you may get the wrong

diagram!!! On screen, diagram is correct.

Diagram for part B

Text for part A

Click print

icon at top

left instead!


Magnetic field

F = ILBsin!

!v

Δq

F = !q v Bsin"=!!q

!t

"(v !t) Bsin"

LI


F = qvBsin!

Force on a charge moving in a magnetic field

Force on a current in a magnetic field

q

!v

!B!

!F


Prob. 21.29: A single turn coil is placed in a uniform magnetic field of 0.25 T. Each side of the coil is of length L = 0.32 m. I = 12 A.

Determine the magnetic force on each side of the coil.

F = 0F = 0

" F

. F

L = 0.32 m

0.32 m

I

L L


Prob. 21.30: The triangular loop of wire carries a current, I = 4.7 A and B = 1.8 T. Find the force acting on each side and the net force.

" F

. FL h

θ = 550

900


A current I flows through the bar. The bar slides down the rails without friction at constant speed.

What is I? In what direction does it flow?

• The component of the weight down the plane must be equal to the component of the magnetic force up the plane...

21.34: B = 0.05 T, verticalm = 0.2 kg = mass of bar

L

I

v constant

m


!v

++

+

– – –

I II

I

Charges inside the moving rod experience a force due to the magnetic field...

Conductor

The moving conductor acts as a generator.

The basis of electromagnetic induction (next chapter)



An alternating current in the voice coil causes the cone of the loudspeaker to move in and out and generate sound – an electrical signal is converted into a sound wave.


Ex. 5: The voice coil of a speaker has a diameter d = 0.025 m, contains 55 turns of wire and is placed in a magnetic field of 0.1 T. The current in the coil is 2 A.

a) Find the magnetic force that acts on the coil and cone.

b) The voice coil and cone have a combined mass of 0.02 kg. Find their acceleration.

m = 0.02 kg

d = 0.025 m

B = 0.1 T


Ions within the water allow a current to flow when a potential difference is applied.

The moving ions feel a force due to the magnetic field that pushes the water out through the back of the boat.

The boat feels an equal and opposite reaction force to the front.

Propulsion with no moving parts.

Magnetohydrodynamics (MHD)


The Torque on a Current-Carrying Coil

L

L

Area of coil, A = Lw

The force on each vertical arm of the coil is: F = ILB

If the coil has N turns: != NIABsin"

w

w

From above! = angle

between normal

to coil and B

The torque is: ! =Fw

2sin" +

Fw

2sin" = I(Lw)B sin" = IAB sin"

(A = Lw)


If the coil has N turns: != NIABsin"

The formula is valid for any shape of coil, not just rectangular.

NIA is known as the “magnetic moment” of the coil.

!= 90!

! = 0o


Variation of torque with φ

!1800 360000 5400

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 100 200 300 400 500 600 700

sin(pi*phi/180)

Torque changes sign! – rotation reverses, no good for an electric motor

Torque: != NIABsin"

Torque, τ


Reverse the current at the right times...

!1800 360000 5400

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 100 200 300 400 500 600 700

sin(pi*phi/180)

Torque, τ

...the torque is always in the same direction and the motor continues to rotate


Electric motor!= 0

!

Commutator reverses direction of current when φ = 0o, 180o...

Torque continues in the same direction

1 2 1

2


Prob. 21.41: The loop is free to rotate about the z-axis.Find the torque on the loop and whether the 35º angle will increase or decrease.

I = 4.4 A

B = 1.8 T

! = 90º – 35º


Ch. 18-22, 24, 25

The next few weeks...


The Torque on a Current-Carrying Coil

L

L

Area of coil, A = Lw

The force on each vertical arm of the coil is: F = ILB

w

w

From above! = angle

between normal

to coil and B

Torque exerted on the coil: ! = NIAB sin "

N = number of turns, A = area of coil


Prob. 21.44: A square coil and a rectangular coil are made from the same length of wire. Each contains a single turn. The long sides in the rectangle are twice as long as the short sides.

Find the ratio of torques the coils experience in the same magnetic field.

• Find the relation between the lengths of the sides of the square and the rectangle, given that the perimeters are of the same length.

• Work out the areas of the two shapes and relate to the torques.

a = 2b

b L

L


Magnetic field produced by currents

Right hand rule #2: the magnetic field wraps around the wire in the direction indicated by the fingers of the right hand.

I


Magnetic field produced by currentsAt a distance r from an infinitely long, straight wire:

B=µ0I

2!r

B

µ0 = “permeability of free space”

= 4!!10"7 T.m/A

I

B

Fingers show direction in which the

magnetic field wraps around

the wire


21.C18: Currents of the same magnitude flow in each of the wires that run perpendicular to the plane of the screen. Choose the direction of the current for each wire, so that when any single current is turned off, the total magnetic field at point P at the centre is directed toward a corner of the square.

X

X

!B1+!B3 = 0

!B4

Currents at opposite corners should be in the same direction

!B1

!B3

!B2

!B2 + !B4 = 0


Prob. 21.49/47: In a lightning bolt, 15 C of charge flows in 1.5 ms. Assuming the lightning bolt can be represented as a long, straight line of current, what is the magnitude of the magnetic field at a distance of 25 m from the bolt?

• What is the current?


Prob. 21.72: Two long, straight wires are separated by 0.12 m. They carry currents of 8 A in opposite directions. Find the magnetic field at A and at B.


Prob. 21.59: The picture shows the end view of three wires that are perpendicular to the screen. The currents in the wires are I1 = I2 = I.

What current I3 is needed in wire 3 to make the magnetic field at the empty corner equal to zero?

I1

I2

I1I1I1

I3

!B1

!B2

!B1+!B2

!B3

L

a

a

a

a

|!B3| = |!B1+!B2|

45º

45º

45º

45º


Force on a charge moving parallel to a conductor

I

Direction in which the magnetic

field wraps around the

wire

B

B =µ0I2!r

Direction of force on the

charge

F = q0vBReplace the moving charge by a current→ attractive forcebetween currents in

same direction

F = q0vB = q0vµ0I

2!r


Magnetic field

F = ILBsin!

!v

Δq

F = !q v Bsin"=!!q

!t

"(v !t) Bsin"

LI


F = qvBsin!

Force on a charge moving in a magnetic field

Force on a current in a magnetic field

q

!v

!B!

!F


Force between long current-carrying wires

Force per unit length of wire 2 is: µ0I1I2

2!d

Equal and opposite force on wire 1 due to current in wire 2

dL

!B1

d

L

!B1

At wire 2: F = I2LB1 = I2Lµ0I1

2!d=µ0I1I2

2!dL

(Due to I1

in wire 1, at position of wire 2)


21.C14: The two outer wires are held in place.

Which way will the middle wire move?

1 2 3

!F12

!F23


Force between current-carrying conductors

Force per unit length of wire:

F =µ0I1I2

2!d

Force is attractive when the currents are in the same direction.

I1 I2

d

x !B1!F!!F

B1 =µoI1

2!d

F = I2LB1


21.C12: A conducting wire is wound into a helical shape. The helix acts as a spring and expands back toward its original shape after the coils are squeezed together and released. The bottom end of the wire just barely touches the mercury (a good electrical conductor) in the cup. After the switch is closed, current in the circuit causes the light to glow. Does the bulb glow continually, glow briefly and then go out, or repeatedly turn on and off like a turn signal on a car?


Prob. 21.54: Find the net force acting on the wire loop that is beside the long, straight wire.

Find the forces acting on wire loop due to current I1.

!F1

!F2

I1 = 12 A

I2 = 25 A

I2 = 25 A

"

"

!B1

!B1

(1)

(2)

B1 is due to I1


Magnetic field at centre of current loop of radius R

B=µ0I

2RN

at centre of a coil of radius R with N turns


Magnetic field at centre of current loop of radius R

B=µ0I

2RN

at centre of a coil of radius R with N turns

N S


Magnetic field produced by currents

B=µ0I

2!r

I

B

B=µ0I

2RN


21.-/53:

At the centre of the loop, B = 0. What is H?

I2 = 6.6 I

1

Due to the loop: B1 =µ0I1

2RDue to the wire: B2 =

µ0I2

2!H

"

!B2!B1

For B1 = B2 :I1

R=

I2

!HSo, H = R

I2

!I1=6.6

!R= 2.1R

Need

B1 = B

2


21.55: Two circular coils are concentric and lie in the same plane. The inner coil contains 140 turns or wire, has a radius r1 = 0.015 m and carries a current I1 = 7.2 A. The outer coil contains 180 turns and had a radius r2 = 0.023 m.

What must be the current in the outer coil so that the net magnetic field at P at the common centre is zero?

r1 = 0.015 m

r2 = 0.023 m

I1 = 7.2 A

I2 = ?

P

N1 = 140

N2 = 180


Magnetic field from a solenoid

n = number of turns per unit length of the solenoidMagnetic field is very uniform at centre of solenoid

B= µ0nI

Inside the solenoid


21.C15:Attractive or repulsive forces?

21.C16:Attractive or repulsive forces?

NS

NS

SN S N

S NSN


Ampère’s Law

Construct a closed path around the two wires.

Split the path into short segments of length !l, measure the magnetic field, B|| in the direction of each

segment.

Add up all of the B|| !l

Then:

I is the net current passing through the surface, I = I1 + I2.

! B||"l = µ0I


Field from a long, straight wire, using Ampère’s Law

Choose a circular path of radius r around the wire.

B is the same all the way around.

so, ! B||"l = B!2#r = µ0I

B=µ0I

2!r

As before!


Prob. 21.61/60: A uniform magnetic field is everywhere perpendicular to the page. A circular path is drawn on the page. Use Ampere’s law to show there can be no net current passing through the circular surface.

!B B|| = 0

"l

! I = 0, zero net current passes through the surface

! B||"l = µ0I around the circle

!B


Prob. 21.62: A very long, hollow cylinder is formed by rolling up a thin sheet of copper. Electric charges flow along the copper sheet parallel to the axis of the cylinder, forming a hollow tube carrying current I.

a) Use Ampere’s law to show that the magnetic field outside the cylinder at a distance r from the axis is:

b) Show that the field is zero inside the cylinder.

B=µ0I

2!r


Prob. 21.34: A thin, uniform rod 0.45 m long and of mass 94 g is attached to the floor by a hinge at point P. A uniform magnetic field B = 0.36 T is directed into the screen. The current in the rod is I = 4.1 A. The rod is in equilibrium. What is the angle θ?

(Hint: the magnetic force can be taken to act at the centre of gravity of the rod).

Floor

mg

90º

F

Centre of gravity

F = ILB

I

L/2

L/2There are forces at the

hinge


Prob. 21.43: Coil: N = 410 turns; area per turn, A = 3.1 " 10-3 m2; current I = 0.26 A. Magnetic field, B = 0.23 T.

A brake shoe is pressed against the shaft to keep the coil from turning. The coefficient of static friction between the shaft and the brake shoe is #s = 0.76. The radius of the shaft is 0.012 m. What is the magnitude of the minimum normal force that the brake shoe exerts on the shaft?

• Find out the maximum torque exerted by the coil

• Equate it to the torque exerted on the shaft by the brake shoe

Brake shoe


Summary of chapter 21

Force on a moving charge: F = qvB sin#

Force on a current-carrying conductor: F = ILB sin#

Torque on a coil: $ = NIAB sin!

Magnetic field from a long, straight wire: B = #o I/2!r

Magnetic field at centre of a circular loop: B = #o NI/2R

Magnetic field at centre of a solenoid: B = #o nI

Ampere’s law: ! B||"l = µ0I

v, I


Chapter 21: Magnetic Forces and Fields Notes/Cha… · 21.5/2: B = 0.3 T, v = 45 m/s, q = 8.4 μC....

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Transcript of Chapter 21: Magnetic Forces and Fields Notes/Cha… · 21.5/2: B = 0.3 T, v = 45 m/s, q = 8.4 μC....