21.4 The Mass Spectrometer

eBmv

qBmvr ==

magnitude of electron charge

eVmv =221

KE=PE

1

The mass spectrum of naturally occurring neon, showing three isotopes.

Application: circular motion of moving ions In a uniform magnetic field: The mass spectrometer

22

2B

Verm

=

eVBmremv

Bm

revm

eBrv

==

=→=

222

221

22

222

2

2

Example: A singly charged positive ion has a mass of 2.5 x 10-26 kg. After being accelerated through a potential difference of 250 V, the ion enters a magnetic field of 0.5 T, in a direction perpendicular to the field. Calculate the radius of the path of the ion in the field.

3

? T 5.0

V 250kg 105.2

C 106.126

19

==

=∆=

=−

−

rB

Vxm

xq

qBmvr

rmvqvBFF cB ===

2

m/s 105.66105.2

)106.1)(250(22 426

19

22

1

×==∆

=

=∆

==∆

−

−

xx

mVqv

qmv

qK

qWV

Example: A singly charged positive ion has a mass of 2.5 x 10-26 kg. After being accelerated through a potential difference of 250 V, the ion enters a magnetic field of 0.5 T, in a direction perpendicular to the field. Calculate the radius of the path of the ion in the field.

We need to solve for the velocity!

m 0.0177)5.0)(106.1(

)568,56)(105.2(19

26

== −

−

xxr

21.3 The Motion of a Charged Particle in a Magnetic Field

Example: Velocity Selector

A velocity selector is a device for measuring the velocity of a charged particle. The device operates by applying electric and magnetic forces to the particle in such a way that these forces balance.

Given B and q , wow should an electric field be applied so that the force it applies to the particle can balance the magnetic force? Solution: by RHR-1: the velocity is to the right (thumb) and the magnetic field (finger) is into the page (“x” marks the tail of an arrow), so the magnetic force on a positive charge is upward (palm up). Here θ = 90° FB = qvB

We therefore need the electric force FE that points down that has the same magnitude: for a positive charge the electric field then needs to point in the same direction as the desired force, and FE = qE.

We want FE = FB qE = qvB

E = vB (and pointing downward) 4

21.5 The Force on a Current in a Magnetic Field

Magnetic force on a current

The magnetic force on the moving charges pushes the wire to the right. Since a current consists of moving charges then a current is subject to the Lorentz force

5

θsinqvBF =

( )

θsinBtvtqF

LI

∆

∆∆

=

θsinILBF =

Magnetic force on a straight current segment of length L

Here θ is the angle between the direction of the current and the magnetic field

In Class Demo

6

Example: A wire carries a current of 22.0 A from west to east. Assume that at this location the magnetic field of Earth is horizontal and directed from south to north and that it has a magnitude of B=0.500×10−4 T. (a) Find the magnitude and direction of the magnetic force on a 36.0 m length of wire. (b) Calculate the gravitational force on the same length of wire if it’s made of copper and

has a cross-sectional area of 2.50×10−6 m2.

21.5 The Force on a Current in a Magnetic Field

7

Example: A wire carries a current of 22.0 A from west to east. Assume that at this location the magnetic field of Earth is horizontal and directed from south to north and that it has a magnitude of B=0.500×10−4 T. (a) Find the magnitude and direction of the magnetic force on a 36.0 m length of wire. (b) Calculate the gravitational force on the same length of wire if it’s made of copper and

has a cross-sectional area of 2.50×10−6 m2.

199 /N 89.7 )m/s kg)(9.81 8050(

kg 8050m) 0.36)(m 1050.2)(kg/m 1094.8()( : wireof Mass

kg/m 1094.8 iscopper of Density (b)(up)N 103.96 T) 10m)(5.00 0.36)(A 0.22(90 Here

field magnetic theandsegment current theof direction thebetween angle theis where sin

upward pointing is force the1- By RHR(a) :Solution

2

2633

33

25-

====

=××=

==×=

×=×==→°=

==

−

−

BG

G

B

B

FF.MgF

.MALvolumeM

ILBF

ILBF

ρρρ

θ

θθ

21.5 The Force on a Current in a Magnetic Field

21.7 Magnetic Fields Produced by Currents

Right-Hand Rule No. 2. Curl the fingers of the right hand into the shape of a half-circle. Point the thumb in the direction of the conventional current, and the tips of the fingers will point in the direction of the magnetic field.

A LONG (∞), STRAIGHT WIRE carrying current produces a magnetic field

rIB o

πµ2

= AmT104 7 ⋅×= −πµo

permeability of free space

Magnitude

Direction

In Class Demo Result to remember: derivation requires integral calculus

8

21.7 Magnetic Fields Produced by Currents Current carrying wires can exert forces on each other.

9

Example: What is the attraction/repulsion force per meter of wire between two infinite straight wires carrying 1.00A each, separated by 1.00 m?

21.7 Magnetic Fields Produced by Currents Current carrying wires can exert forces on each other.

10

Example: What is the attraction/repulsion force per meter of wire between two infinite straight wires carrying 1.00A each, separated by 1.00 m?

mN1000.2

sC

mCsN1000.2

T)1000.2(A) 00.1( 2

90 ,sin

T1000.2)m 00.1(2

A) 00.1)(m/AT 104( 2

77

7

21012

2

21

12221

77

101

−−

−

−−

×=⋅⋅⋅

×=

×=

==

°==

×=⋅×

==

rIIBI

LF

BLIFr

IB

πµ

θθπ

ππ

µ

This is in fact the original SI definition of an ampere (A).

1 m

21.7 Magnetic Fields Produced by Currents

SOLENOID

11

Result to remember: derivation requires integral calculus

Magnetic field in the interior of a solenoid Is approximately uniform L

NInIB oo

µµ ==

number of turns per unit length n = N / L

Magnitude In the limit of infinite length, but finite n

21.7 Magnetic Fields Produced by Currents

12

R

A CIRCULAR LOOP OF WIRE

At the center of the circular loop only

R.H. B

Two different applications of RHR-2 for the same field configuration

Result to remember: derivation requires integral calculus

RIB o

2µ

=Magnitude

21.7 Magnetic Fields Produced by Currents Example: Net Magnetic Field

A long straight wire carries a current of 8.0 A and a circular loop of wire carries a current of 2.0 A and has a radius of 0.030 m. Find the magnitude and direction of the magnetic field at the center of the loop.

13

21.7 Magnetic Fields Produced by Currents

Example: Net Magnetic Field

A long straight wire carries a current of 8.0 A and a circular loop of wire carries a current of 2.0 A and has a radius of 0.030 m. Find the magnitude and direction of the magnetic field at the center of the loop.

14

−=−=

RI

rI

RI

rIB ooo 2121

222 πµµ

πµ

( )( ) T101.1

m 030.0A 0.2

m 030.0A 0.8

2AmT104 5

7−

−

×=

−

⋅×=

ππB

21.7 Magnetic Fields Produced by Currents The field lines around a current loop resemble those around the bar magnet.

15

Attraction

Repulsion

21.6 The Torque on a Current-Carrying Coil

16

Put a rectangular loop of current I and length (height) L, and width w in a uniform magnetic field B. The loop is mounted such that it is free to rotate about a vertical axis through its center. We will consider the forces on each segment and the resulting torque from each. Using RHR-1: The force on segment 3 points down, and that on segment 4 points up. F3 and F4 are also equal in magnitude and cancel one another. The magnitudes F3 = F4 = IwBsin(90°-φ) = IwBcosφ also change with the rotation angle φ But both F3 and F4 are directed parallel to the axis, and results in no torque.

φ is the angle between the “normal” to the loop and the magnetic field

3

4

3

Top view

21.6 The Torque on a Current-Carrying Coil

17

Looking at segments 1 and 2 which have the current running vertically. By RHR-1, force F1 on segment 1 (current up) points into the page, for all values of φ. Also by RHR-1, force F2 on segment 1 (current down) points out of the page. They cancel each other to yield no net force on the loop. However, F1 and F2 both tend to turn the loop in the clockwise sense (as seen in the top view). The torques from the two forces are each

1

2

φ is the angle between the “normal” to the loop and the magnetic field

Top view

( ) φτ sin22,12,1

=

wF

φτττ sin21 Fw=+=°=

==90 since

sinθ

θ ILBILBF

21.6 The Torque on a Current-Carrying Coil

18

Top view

φτττ sin21 Fw=+=

The torque τ is maximum when the normal of the loop is perpendicular to the magnetic field, and zero when the normal is parallel to the field.

The torque tends to cause the loop normal to become aligned to the field, just like on a bar magnet. Current loop = magnetic dipole

( ) φφτ sinsinNet torque IABwILB ===°===

90 sincesin

θθ ILBILBF

loop ofarea == LwA

φτ sinmomentmagnetic

BNIA=number of turns of wire

m

21.6 The Torque on a Current-Carrying Coil

19

Example The Torque Exerted on a Current-Carrying Coil A coil of wire has an area of 2.0x10-4m2, consists of 100 loops or turns, and contains a

current of 0.045 A. The coil is placed in a uniform magnetic field of magnitude 0.15 T.

(a) Determine the magnetic moment of the coil. (b) Find the maximum torque that the magnetic field can exert on the coil.

21.6 The Torque on a Current-Carrying Coil

Example The Torque Exerted on a Current-Carrying Coil A coil of wire has an area of 2.0x10-4m2, consists of 100 loops or turns, and

contains a current of 0.045 A. The coil is placed in a uniform magnetic field of magnitude 0.15 T.

(a) Determine the magnetic moment of the coil. (b) Find the maximum torque that the magnetic field can exert on the coil.

(a)

( )( )( ) 2424

momentmagnetic

mA100.9m100.2A 045.0100 ⋅×=×== −−NIAm

( )( ) mN104.190sinT 15.0mA100.9sin 424momentmagnetic

⋅×=⋅×== −− φτ BNIA(b)

20

21.6 The Torque on a Current-Carrying Coil Application: The basic components of a dc motor.

21

The brushes switches the direction of the current so that the torque is always in the same direction continuous rotation

21.8 Ampere’s Law

AMPERE’S LAW FOR STATIC MAGNETIC FIELDS For any current geometry that produces a magnetic field that does not change in time,

IB oµ=∆∑ ||

net current passing through surface bounded by path

22

“positive” sense for the current bounded is related to the direction of the path traveled by RHR-2

Positive direction for bound currents

21.8 Ampere’s Law Example An Infinitely Long, Straight, Current-Carrying Wire

Use Ampere’s law to obtain the magnetic field.

23

21.8 Ampere’s Law

Example An Infinitely Long, Straight, Current-Carrying Wire

Use Ampere’s law to obtain the magnetic field.

IB oµ=∆∑ ||

( ) IB oµ=∆∑

IrB oµπ =)2(

rIB o

πµ2

=

24

1. By symmetry, the magnetic field must have cylindrical symmetry: depends only on distance r to the wire.

2. The magnetic field should be generally perpendicular to the current (by Right Hand Rule).

Choose a circular path (at constant r) to match the symmetry The magnetic field lines circulate around the line by RHR-2: traverse path in the CCW direction as seen from the top The field is parallel to the path and has constant magnitude on the path.