Problem 27.15

Fig. 27.46

An electron at point A has a speed

of 1.41 x 106 m/s. Find

(a) the magnitude and direction of

the magnetic field that will cause the

electron to follow the semicircular path

from A to B, and

(b) the time required for the electron to

move from A to B.

PHYS 153 08W 1PHYS 153 08W 1PHYS 153 08W 1

Fig. 27.46

Problem 27.31 Determining the mass of an isotope.

The electric field between the plates of the velocity selector in the

Bainbridge mass spectrometer (Fig. 27.24) is 1.12 x 105 V/m, and

the magnetic field is 0.540 T. A stream of singly charged selenium

ions moves in a circular path with a radius of 31.0 cm in the magnetic

field. Determine the mass of one selenium ion and the mass number

of the selenium isotope. (The mass number is equal to the mass of

the isotope in atomic mass units, rounded to the nearest integer. One

atomic mass unit = 1u = 1.66 x 10-27 kg).

PHYS 153 08W 2PHYS 153 08W 2PHYS 153 08W 2Fig. 27.24

The force exerted by a magnetic field.

x x x x x x x

x x x x x x x

x x x x x x x •

positive charge entering a magnetic field

The force on a single charge

is given by

→→→

= BxvqF

When a wire carries a current, in a

magnetic field, there is a force on the

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magnetic field, there is a force on the

wire that is equal to the sum of the

magnetic forces on the charged particles

whose motion produces the current.

The number of charges in length is

, where n is the density and is the

volume.

Fig. 27.25

nAlBxvqF d )(→→→

=∴

But the current

lnAl Al

AnqvI d=

Hence the force can be written→→→

= BxlIF

Note: Current is not a vector. is. The direction in which current

flows is given by , or .

I→

l I→

l→

ld

D.C. motor

In everyday life, there are countless examples of motors. It is important to

understand how they work. First though, we have to learn about -

Forces and torques on a current loop.B

(valid for both + and – charges)

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Forces and torques on a current loop.B

I

l

B

φ

I

l

Recall for a current carrying wire

in a magnetic field,

φsinIlBBxlIF ==→→→

IlBF = when o90=φThe diagrams illustrate these two cases.

To understand this is important when

studying the case of torques on a current

carrying loop of wire in a B field.

Fig. 27.31Fig. 27.31

This figure shows a current

carrying loop of wire in a

magnetic field.

A normal to the plane of the

loop makes an angle to

the magnetic field, B.

Consider the sides of length b.

The force on each side is

φ

φφ cos)90sin(' IbBIbBF =−=

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Fig. 27.31φφ cos)90sin( IbBIbBF =−=

They are designated and ∧

jF ' )('∧

− jF

and are equal and opposite. There is no net force in the direction.

Consider the sides of length a. The force on each side is

IaBF =

These forces are designated and ∧

iF )(∧

− iF

and are equal and opposite. There is no net force in the direction.

y±

x±

But there is a torque acting on the sides of length a.

φφφτ sin)sin)((sin)2

(2 IABbIaBbF ===

Torque is a maximum when (Fig. 27.31b), and zero when , or o90=φ o0=φ o180

(Fig. 27.31c).

We can write torque as

φµτ sinB= where (or where is the number of turns of wire) IA=µ NIA

is the magnetic dipole moment of the loop.

N

→

PHYS 153 08W 6PHYS 153 08W 6PHYS 153 08W 6

Since is a vector, is a vector and the direction is given by the R.H. rule:

Place the fingers of the R.H. in the direction of and the thumb points in

→

A→

µI

→

µthe direction of (when it is perpendicular to the plane of the loop). →→→

=∴ Bxµτ

(recall the similarity to the torque exerted by an electric field on an electric dipole moment,→→→

= Expτ ).

Using this symmetry and recalling that the potential energy of an electric

dipole is , can make a reasonable guess that the potential →→

•−= EpU

energy for a magnetic dipole moment is

→→

•−= BU µ

U Is greatest when and are antiparallel, and least when they are

→

µ→

B

parallel. when and are perpendicular to each other. 0=U→

µ→

B

PHYS 153 08W 7PHYS 153 08W 7PHYS 153 08W 7

parallel. when and are perpendicular to each other. 0=U µ B

Note that the rectangular current loop is not a special case. The relationships

above are valid for a plane loop of any shape. The

irregular shape to the right can be approximated by

an infinite number of rectangles. The currents in

adjacent sides cancel and that leaves just the

current in the outer boundary. The results above

are valid for each of the rectangles, and hence

valid for the irregular shape. Fig. 27.33

Question: What is the torque on a solenoid

situated in a magnetic field as shown?

For a single loop,

φµφτ sinsin BIAB ==A solenoid is equivalent to N circular

loops, side by side. So in this case,

NIA=µThe torque as usual is given by

φµτ sinB=

PHYS 153 08W 8PHYS 153 08W 8

Fig. 27.34

φµτ sinB= , so for the solenoid,

φτ sinNIAB=

Now we are ready to discuss the D-C motor.

The above ideas of current carrying loops and torques are applicable in this

case.

PHYS 153 08W 9PHYS 153 08W 9

Fig. 27.39

A current carrying loop lies in a magnetic field generated by a permanent

magnet. The battery is connected to the coil through commutators which

keep the current flowing in the same direction in the loop as it turns.

In (a), note the orientation of with respect to . The torque will turn

the loop counter-clockwise.

→

µ→

B

In (b), each brush which touches the commutators touches both

commutators simultaneously.

Hence, through the coil, , and .0=I 0=→

µ 0=→

τMomentum keeps the loop turning.

In (c), the coil has turned with respect to (a), but because of the o180→

µcommutators, is in the same direction as it was in (a) and the torque

will continue to turn the loop counter-clockwise.

The coil will turn at top speed when the turning torque is balanced by a

resisting torque due to air resistance and friction.

For N turns of wire in the loop, the torque will be N times stronger.

PHYS 153 08W 10PHYS 153 08W 10

Question:

What improvements could be made to this simple motor set-up to make

it more efficient?

Back emfs

When a wire moves in a magnetic field, a current is induced in it to

oppose the motion causing it. In the motor discussed above, the induced

current is in the direction opposite to the current shown. This induced emf

is called a back emf. We will study this in more detail later.

Hall Effect

The Hall effect is the classic way of

determining the sign of the mobile

charge carriers in a material, and the

density of current carrying charges in

the material.

A current flows to the right in the +x

direction. For a metal conductor, the

drift velocity, vd, of the charge carriers

(electrons) is to the left (diagram (a)).

PHYS 153 08W 11

Fig. 27.41

(electrons) is to the left (diagram (a)).

For a semiconductor, for example, the

drift velocity of the charge carriers (holes)

is to the right (diagram (b)).

In both cases, the magnetic field produces

a force on the moving charges in the +z

direction.

In (a), an excess of negative charges accumulates near the top edge of the

conductor. In (b), the charge accumulating along the top is positive. The net

result of this is to create an electric field to oppose the motion of the charge,

negative charge in (a), and positive charge in (b).

The transverse potential difference is called the Hall voltage.

Eventually, the electrostatic force balances the magnetic force.

0)()( =+−∧∧

kBqvkqE d, and BvE d−=

(when is negative, is positive, and vice versa).dv E

PHYS 153 08W 12

dv E

From before, the current density dx nqvJ =

Eliminate from these last two equation and obtain,dv

E

BJnq

yx−=

xJ and yBare positive. If is positive (diagram (a)), then nq is negative,E

and vice versa (diagram (b)).

(Hall effect)

Since the magnitude of q is known (equal to the electron charge), then n can

be calculated.

32910~ −mnFor a metal conductor,

How magnets work

B

What will happen to a bar magnet when it is

placed in a magnetic field, and at some angle

to it?

It will rotate until it is aligned with the field. This

PHYS 153 08W 13

bar magnet

It will rotate until it is aligned with the field. This

means that the bar magnet has a magnetic

dipole moment. The origin of this dipole moment

is the electron which spins and acts somewhat

like a current loop (look back on slide 5). An

electron has a magnetic moment. In a

permanent magnet, these dipoles are well

aligned. The magnetic dipole moment for a

permanent magnet points from S to N.

S N

→

µB

Fig. 27.37

In an unmagnetized object (e.g. iron in diagram (a)), there is no overall alignment

of the magnetic moments, and the vector sum is zero. In the presence of a

magnet, the magnetic moments align with the B field and the object acquires a

net magnetic moment (diagram (b)). This new magnetic moment of the object

PHYS 153 08W 14

Fig. 27.37

(e.g. the magnet above, or an iron nail) is the reason why the object is attracted

by the B field of the magnet.

Attraction and repulsion

Fig. 27.36→

µ→

µ

In Fig. 27.36 (a), the current loop has a magnetic dipole moment which is

anti-aligned with the magnetic moment of the bar magnet. The force on a

section of the loop has a radial component, and a component to the right.

The radial components cancel and the net force on the loop is to the right.

So, when the magnetic moments are anti-aligned, the loop is repelled from

the bar magnet.

In (b), the current loop has a magnetic dipole moment which is aligned with

that of the bar magnet. The net force on the current loop is now to the left.

So, when the magnetic moments are aligned, the loop is attracted to the bar

magnet.

PHYS 153 08W 15

For a class of materials which are called ferromagnetic (iron, nickel, cobalt

and alloys of these) which are unmagnetized, the atomic magnetic moments

tend to align with the B field when in the presence of a permanent magnet.

The aligned dipole moments are represented by (b) in Fig. 27.36, and the

object is attracted to the magnet. Also (b) explains the attraction between

two bar magnets end to end, with the S pole of one next to the N pole of the

other.

Diagram (a) in Fig. 27.36 explains the repulsion between two bar magnets

when similar poles are put next to each other.

Problem 27.51

The figure shows a portion of a

silver ribbon with z1=11.8 mm and

y1= 0.23 mm, carrying a current of

120 A in the +x direction. The ribbon

lies in a uniform magnetic field, in the

y direction, with magnitude 0.95 T.

If there are 5.85 x 1028 free electrons

per cubic metre, use the simplified model

of the Hall effect to find

PHYS 153 08W 16

of the Hall effect to find

(a) the magnitude of the drift velocity of the

electrons in the x direction,

(b) the magnitude and direction of the electric field in the z direction

due to the Hall effect,

(c) the Hall emf.