32N 28 Sources of Magnetic Field

Sources of Magnetic Fields

Ms. Arlyn D. Macasero Chap 28 - Sources of Magnetic

Fields

1

Magnetic Field of Moving Charge


Fields

2

field point

+

r

r̂

v

P

q

source point

0

2

ˆ

4

q

r

rB

v

B

0

2

sin

4

qB

r

v

Magnetic permeability of

vacuum:

0 = 4107 Tm/A

The magnetic field at point P

due to the moving charge:

Magnetic Field of Moving Charge


Fields

3

Magnetic field lines form circles centered around the line of motion

of the charge (using the right-hand rule)

0

2

ˆ

4

q

r

rB

v 0

2

sin

4

qB

r

v

Example 1. Forces between two moving charges


Fields

4

A pair of point charges q1 = +5.00C and q2 = 3.00C are moving at

speeds v1 = 7.50x104m/s and v2 = 3.20x104m/s as shown in the

figure. (a) What is the magnitude and direction of the magnetic force

that q1 exerts on q2? (b) What is the net magnetic field (magnitude and

direction) at the origin?

Ox

y

+

-

q1

q2

v1

v20.30 m

0.40 m

0

2

sin

4

qB

r

v



Fields

5

A pair of point charges q1 = +5.00C and q2 = 3.00C are moving at

speeds v1 = 7.50x104m/s and v2 = 3.20x104m/s as shown in the

figure. (a) What is the magnitude and direction of the magnetic force

that q1 exerts on q2? (b) What is the net magnetic field (magnitude and

direction) at the origin?

Answer:

B21 = 9.0 x 10-8 T (-z – axis )

F21 = 8.6 x 10-9 N (+x – axis )

BNet = 3.6 x 10-7 T (-z – axis )Ox

y

+

-

q1

q2

v1

v20.30 m

0.40 m

0

2

sin

4

qB

r

v

Magnetic Field of a Current Element


Fields

6

Principle of Superposition of Magnetic Fields

The total magnetic field caused by several moving charges is the

vector sum of the fields caused by the individual charges.

Law of Biot and Savart

Magnetic field lines form

circles centered around the

direction of the current

(using the right hand rule)

Magnetic Field of a Current Element


Fields

7

Principle of Superposition of Magnetic Fields

The total magnetic field caused by several moving charges is the

vector sum of the fields caused by the individual charges.

0

2

ˆ

4

Idd

r

l rB

Magnetic Field of Straight Current-Carrying Conductor


Fields

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0

2

sin

4

IdldB

r

Magnetic field at P due to

current element:



Fields

9

0

2

sin

4

IdldB

r

Magnetic field at P due to

current element:

0

2 2 3 24 ( )

I xdB dy

x y

0

2 2

2

4

I aB

x x a

In terms of x and y:

The total magnetic field at P:

In the limit as a, this becomes

0

2

IB

x

Or, for any radial distance r around

the conductor

0

2

IB

r



Fields

10

The magnetic field lines around a long straight current-carrying

conductor form circles centered around the current (using the right-

hand rule)

Example 2. Magnetic Field around a Straight Wire


Fields

11

You want to produce a magnetic field of magnitude 1.00 mT at a

distance of 5.00 cm from a long, straight wire. (a) What current is

required to produce this field? (b) With the current obtained in (a), what

is the magnitude of the field at a distance of 10.0 cm from the wire and

at 20.0 cm?

Example 2. Magnetic Field around a Straight Wire


Fields

12

You want to produce a magnetic field of magnitude 1.00 mT at a

distance of 5.00 cm from a long, straight wire. (a) What current is

required to produce this field? (b) With the current obtained in (a), what

is the magnitude of the field at a distance of 10.0 cm from the wire and

at 20.0 cm?

Answer:

(a) The required current can be solved from

(b) At a distance 10.0 cm from the wire,

and at 20.0 cm from the wire,

3

0

7

0

2 2 (0.05m)(1.00 10 T) 250 A

2 4 10 T m/A

I rBB I

r

0 (250 A)0.500 mT

2 (0.1 m)B

0 (250 A)0.250 mT

2 (0.2 m)B

Ampere’s Law


Fields

13

Recall, that we can find the

net electric field due to any

distribution of charges using

Gauss’s Law for electric fields.

enc

0

qd

E A

Ampere’s Law


Fields

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Recall, that we can find the

net electric field due to any

distribution of charges using

Gauss’s Law for electric fields.

Similarly, we can find the

net magnetic field due to

any distribution of current

using Ampere’s law

?

B

enc

0

qd

E A

Ampere’s Law


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where the integration is around a closed loop and

the current ienc is the net current encircled by that closed loop.

0 encd i B s

The line integral of the magnetic field around any closed path is

proportional to the net current that pierces the loop.

Ampere’s Law


Fields

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where the integration is around a closed loop and

the current ienc is the net current encircled by that closed loop.

Curl your right hand around the closed loop with the fingers pointing inthe direction of integration. A current through the loop in the generaldirection of your outstretched thumb is assigned a plus sign, and acurrent generally in the opposite direction is assigned a minus sign.

0 encd i B s

The line integral of the magnetic field around any closed path is

proportional to the net current that pierces the loop.

Ampere’s Law on an Infinitely Long, Straight

Current-Carrying Wire


Fields

17

Use Ampere’s law to obtain the magnetic field produced by the

current in an infinitely long, straight wire.

0 enc

cos

d i

Bds

B s

1

0

0

0

0

(2 )

2

I

B ds I

B r I

IB

r

ds

ds

Magnetic Field Inside a Long Straight Wire with

Uniformly Distributed Current


Fields

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path of

integration

Since, in a straight wire conductor

𝑩 ∙ 𝒅𝒔 = 𝑩 𝒅𝒔 = 𝑩 𝟐𝝅𝒓

𝑖𝑒𝑛𝑐 = 𝐽 𝜋𝑟2 =

𝐼

𝜋𝑅2𝜋𝑟2

𝐵 2𝜋𝑟 = 𝜇𝑜𝑖𝜋𝑟2

𝜋𝑅2

𝐵 2𝜋𝑟 = 𝜇𝑜𝐼 ⟹ 𝐵 =𝜇𝑜𝐼

2𝜋𝑟

𝑩 =𝝁𝒐𝒊

𝟐𝝅𝑹𝟐𝒓 (inside straight wire)


Fields

19

Magnetic Field of a Solenoid

For a long ideal solenoid 𝑩 ∙ 𝒅𝒔 = 𝝁𝒐𝒊𝒆𝒏𝒄

𝑖𝑒𝑛𝑐 = 𝑖(𝑛ℎ)

𝐵ℎ = 𝜇𝑜𝑖𝑛ℎ

𝑩 = 𝝁𝒐𝒊𝒏 (ideal solenoid)


Fields

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Magnetic Field of a Toroidal Solenoid

The figure shows a toroidal solenoid,

which we may describe as a solenoid bent

into a torus (doughnut). From the

symmetry we see the field forms

concentric circles inside the toroid as

shown in figure (b).

path of integration

𝐵 2𝜋𝑟 = 𝜇𝑜𝑖𝑁

𝑩 =𝝁𝒐𝒊𝑵

𝟐𝝅

𝟏

𝒓

Example 3. Magnetic Field of in a Solenoid


Fields

21

A solenoid of length 20.0 cm and radius 3.00 cm is tightly

wound with 500 turns of wire. The current in the windings is

6.00 A. What is the magnitude of the magnetic field at the

center of the solenoid?

Example 3. Magnetic Field of in a Solenoid


Fields

22

A solenoid of length 20.0 cm and radius 3.00 cm is tightly

wound with 500 turns of wire. The current in the windings is

6.00 A. What is the magnitude of the magnetic field at the

center of the solenoid?

Answer:

The magnetic field within the solenoid is

7

00

4 10 (6.00)(500)18.8 mT

0.2

INB In

L



Fields

23

Two positive point charges q1 = q2 =

1.00C move with equal and opposite

velocities where v = 3.00106 m/s, as

shown in the figure. When the protons

are 1.00 m apart, an observer at point P

midway between the particles measures

the forces on each. What is the

magnitude and direction of the force that

each proton exerts on the other?

+

+

P1.00 m

v1

v2

q1

q2

B12F12

B21

F21



Fields

24

Two positive point charges q1 = q2 =

1.00C move with equal and opposite

velocities where v = 3.00106 m/s, as

shown in the figure. When the protons

are 1.00 m apart, an observer at point P

midway between the particles measures

the forces on each. What is the

magnitude and direction of the force that

each proton exerts on the other?

+

+

P1.00 m

v1

v2

q1

q2

B12F12

B21

F21

Answer:

Magnitude: F12 = F21 = 9.0 x 10-7 N

Example 5. Forces on Current-Carrying Conductors


Fields

25

The long straight wire AB in the figure carries a current of 14.0A. The

rectangular loop whose long sides are parallel to the wire carries a

current of 5.00A. Find the magnitude and direction of the net force

exerted on the loop by the magnetic field of the wire.

20.0

cm

10.0 cm

2.6 cm

14.0 A5.00 A

B

A

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32N 28 Sources of Magnetic Field