03 Force on Moving Charge.pptx

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(2)

Magnetic

Field Ve

loci ty

(3)

Direction of Magnetic Force

The right hand rule:

With a flat

right

hand,

point

thumb

in direction

of velocity v,

fingers

in

direction of

B

field. The

flat

hand

pushes in the

direction of

force F

.

The force is greatest when the velocity

v

is

perpendicular to the B field. The deflection

decreases to zero for parallel motion.

The force is greatest when the velocity

v

is

(4)

+

(5)

(6)

(7)

(8)

X X X X X X X X X X X X X X X X

++

+

++

(9)

+

(10)

+

(11)

+

(12)

F ∝

q·v·B

1

Newton

1

Coulomb

·1

·

B

meter second

B = 1Tesla

(13)

Magnetic Force on Moving Charge

Imagine a tube that

projects charge +q

with velocity v into

perpendicular B field.

Magnetic force F on charge moving in B field.

Experiment shows:

Each of the following results in a greater magnetic

force F

: an increase in

velocity

v, an increase in

charge

q, and a larger

magnetic field B

.

(14)

+

Greatest

force

when

moving

perpendicular

to field.

+

No

force

when

(15)

+

q

+

q

sin

(16)

Definition of B-field

Experimental observations show the following:

By choosing appropriate units for the constant of proportionality, we can now define the B-field as:

A magnetic field intensity of one tesla (T) exists in a region of space where a charge of one coulomb (C) moving at 1 m/s perpendicular to the B-field will

experience a force of one newton (N).

A magnetic field intensity of one tesla (T) exists in a region of space where a charge of one coulomb (C)

moving at 1 m/s perpendicular to the B-field will experience a force of one newton (N).

sin or

constant

sin

F

qv

q

(17)

Example 1. A

2-nC

charge is projected with velocity

5 x 10

4

m/s

at an angle of

30

0

with a

3 mT

magnetic field as shown.

What is the magnitude and direction of the resulting force?

v sin f 300

v

B

v

F

Draw a rough sketch. q = 2 x 10-9 C

v = 5 x 104 m/s

B = 3 x 10-3 T

q = 300

Using right-hand rule, the force is seen to be upward.

Resultant Magnetic Force: F = 1.50 x 10-7 N, upward

B

-9 4 -3 0

sin

(2 x 10 C)(5 x 10 m/s)(3 x 10 T)sin 30

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Forces on Negative Charges

Forces on negative charges are opposite to those on positive charges. The force on the negative charge requires a left-hand rule to show downward force F.

N

S

N

S

B

v

F

Right-hand rule for

positive q

F

B

v

Left-hand

(19)

x x x x

. . . .

FORCE

C

U

R

E

N

T

(20)

x

FORCE

C

U

R

E

N

T

Mag FIELD

(21)

. . . .

FORCE

C

U

R

E

N

T

Mag FIELD

(22)

+ + + + + + + + + + + + + +

-E

+

· · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · ·

B

+

Force

is always down

regardless of

motion

!

Force

is always

perpendicular to

motion

!

(23)

+

B

E

Cu

rr

en

t

C

u

rr

en

t

+

(24)

Crossed E and B Fields

The motion of charged particles, such as electrons, can be controlled by combined electric and magnetic fields.

x x x x x x x

x +

-e

-v

Note: F_E on electron is upward and

opposite E-field.

But, F_B on electron is

down (left-hand rule). Zero deflection

when F_B = F_E

B

v

F

_E

E e

-

-B

v

F

_B

(25)

-The Velocity Selector

This device uses crossed fields to select only those velocities for which F_B = F_E. (Verify directions for +q)

x x x x x x x

x +

-+q

v Source

of +q

Velocity selector When F_B = F_E:

By adjusting the E and/or B-fields, a person can select only those ions with the desired velocity.

qvB qE



E

v

B

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Example 2. A lithium ion,

q

= +1.6 x 10

-16

C

, is projected

through a velocity selector where

B = 20 mT

. The E-field is

adjusted to select a velocity of

1.5 x 10

6

m/s

.

What is the electric field E?

x x x x x x x

x +

-+q

v Source

of +q

V

E = vB

E = (1.5 x 106 m/s)(20 x 10-3 T);

E = 3.00 x 104 V/m

E

v

B

(27)

Circular Motion in B-field

The magnetic force F on a moving charge is always perpendicular to its velocity v. Thus, a charge moving in a B-field will experience a centripetal force.

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

+

+ +

+ Centripetal F_c = F_B

R

F_c

The radius of path is:

2

;

C B

mv

F

qvB

R



2

mv

qvB

R



C B

F



F

mv

R

qB

(28)

Mass Spectrometer

+q

R

+

-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 Photographic plate m₁ m₂ slit

Ions passed through a velocity selector at

known velocity emerge into a magnetic field as shown. The radius is:

The mass is found by measuring the radius R:

(29)

Example 3. A Neon ion,

q = 1.6 x 10

-19

C

, follows a path of

radius

7.28 cm

. Upper and lower

B = 0.5 T

and

E = 1000 V/m

.

What is its mass?

v = 2000 m/s

m = 2.91 x 10-24 kg

+q

R

+

-

x x _{x x}

x x x x

Photographic plate

m

slit 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

mv

R

qB



m

qBR

v



1000 V/m

0.5 T

E

v

B



-19

(30)

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Vector Cross Product

(32)

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Vector Cross Product

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Blue

cross

Green

=

Red

http://mathinsight.org/cross_product

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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

F

(36)

l

t

(37)

(38)

+

-B

B

0 1 2 3 4 5 6 7

∝

0

2

I

B

r







0

2

I

B

r





(39)

+

-+

-Where will the magnitude

of the B-field equal zero?

A

B

C

D – There is no place where the B-field will equal be zero

Current

(40)

+

-+

-Where will the magnitude

of the B-field equal zero?

A

C

B _D _{– There is no place}

where the B-field will equal be zero

Current

(41)

.

+

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

(42)

I can . . .

• describe the magnitude and direction of the force on a charge moving through a magnetic field.

• describe the magnitude and direction of the force on a wire with current moving through a magnetic field.

• relate the strength of the magnetic field around a wire to the distance from the wire.