Electromagnetism and magnetic circuit 6

8/9/2019 Electromagnetism and magnetic circuit 6

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From the equation for the magnetic field of a moving chargedparticle, it is easy to show that a current I in a little length dl

of wire gives rise to a little bit of magnetic field.

dB

r

r

dl

The Biot-Savart Law

4

vr

r l0

2

I d rdB =

r

I

You may see the equation written using rr =r r .

16-Apr-10 2Chetan Upadhyay


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Applying the Biot-Savart Law

I

ds

r

r

U

dB

4

vr rr

0

2

I s r rr r

r r

4

0

2I s si

r

r rB = dB

Homework Hint: if you have a tiny piece of a wire, just calculate dB; no need to integrate.16-Apr-10 3Chetan Upadhyay


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Example: calculate the magnetic field at point P due to a thinstraight wire of length L carrying a current I. (P is on the

perpendicular bisector of the wire at distance a.)

4

vr

r0

2

I ds rdB =

r

vrds r ds si k

4

0

2

I ds sidB

r

ds is a i fi itesimal qua tity i the directio ofdx, so

4

0

2

I dx sidB

r

I

y

r

x

dBP

ds

Ur

x

z

a

L



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2 2r = x +aa

sin =

r

I

y

r

x

dBP

ds

Ur

x

z

4

0

2

I dx sindB =

r

a

4 4

0 0

3/232 2

I dx I dxdB = =

r x

4L/2

0

3/2-L/2 2 2

I dx aB =

x +a

4 L/2

0

3/2-L/2 2 2

I dxB =

x

L



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I

y

r

x

dBP

ds

Ur

x

z

a

4 L/2

0

3/2-L/2 2 2

I a dxB =

x +a

look integral up in tables, use theweb,or use trig substitutions

3/2 1/2

2 2 2 2 2

dx x=

x +a a x +a

4

L/2

0

1/22 2 2

-L/2

I xB =

ax

a

4

-

0

1/2 1/22 22 2 2 2

I a L/2 -L/2=

a L /2 +a a -L /2 +a

L



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I

y

r

x

dBP

ds

Ur

x

z

a

4

-

0

1/22 2 2

I a 2L/2B =

a L /4 + a

4

0

1/22 2

I L 1B =

a L /4 + a

0

2 2

I L 1B =

2 a L + 4a

0

2

2

I 1B =

2 a 4a1+

L



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I

y

r

x

dBP

ds

Ur

x

z

a

0

2

2

I 1B =

2 a 4a1+

L

When Lpg,

.

0I

B =2 a

0I

B =2 r

or The r in this equation has a differentmeaning than the r in the diagram!



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Example: Long StraightConductor

What is the

magnetic field 4mm

from a long straight

conducting wire

carrying a 5 amp

current?

T

rIB

4

3

7

0

105.21042

5104

2

v! vv

vv!

!

T

TT

Q


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I

B

r

Magnetic Field of a Long Straight Wire

Weve just derived the equation for the magneticfield around a long, straight wire*

0IB =

2 r

with a direction given by a new right-handrule.

*Dont use this equation unless you have a long, straight wire!



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Looking down along the wire:I

B

The magnetic field is notconstant.

At a fixed distance r from the wire, the magnitude of themagnetic field is constant.

The magnetic field direction is always tangent to the

imaginary circles drawn around the wire, and perpendicularto the radius connecting the wire and the point where thefield is being calculated.



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Magnetic Field of a Current-Carrying Wire

It is experimentally observed that parallel wires exert forces oneach other when current flows.

I1 I2

F12 F21

I1 I2

F12 F21



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The magnitude of the force depends onthe two currents, the length of the wires,and the distance between them.

0 1 2I I L

F = 2 d

I1 I2

F12 F21

d

L

The wires are electrically neutral,so this is not a Coulomb force.

We showed that a long straight wire carrying a currentI gives rise to a magnetic field B at a distance r from

the wire given by

0IB =2 r

I

B

rThe magnetic field of one wire exerts a force on anearby current-carrying wire.

This is NOT a

starting equation



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Example: use the expression for B due to a current-carryingwire to calculate the force between two current-carrying wires.

I1 I2

F12

d

L

0 22

IB = k

2 d

vr r r

12 1 1 2F =I L B

L2L1

B2

vr

0 212 1

I= I Lj k

2 d

x

y

r0 1 2

12

I I L

= i2 d

The force per unit length of wire is

r

0 1 212 I I= i.L 2 d



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I1 I2

F12 F21

d

L

r0 1

1

IB = - k

2 d

vr r r

21 2 2 1F =I L B

L2L1

B18

v

r0 1

21 2

IF = I Lj k

2 d

x

y

r0 1 2

21

I I L

F =- i2 d

The force per unit length of wire is

r

0 1 221 I IF = - i.L 2 d



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If the currents in the wires are in the opposite direction, theforce is repulsive.

I1 I2

F12 F21

d

L

L2L1y



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Force between two parallel

current-carrying straight wires (1)

1. Parallel wires with current flowing

in the same direction, attracteach other.

2. Parallel wires with current flowing

in the opposite direction, repel


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Force between two parallel

current-carrying straight wires (2)

Note that the force exerted on I2 by I1is equal but opposite to the force

exerted on I b I .

a

II

F

o

T

Q

2

21 N!


20/30

Problem 1

Two parallel conductors of 20 mm

diameter each carrying 120 A in opposite

directions are separated by an air space of

60 mm. The conductors are 10 m long.

Find the force on each conductor.

16-Apr-10 Chetan Upadhyay 20

F = 0.36 N

vv

--1 2

12 21 1 2

4 10 I I L LF =F = =2 10 I I

2 d d


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Problem 2

Force between two wires carrying

currents in opposite direction is 20.4 kg/m.

When they are placed parallel with their

axis 5 cm apart. Calculate the current in

one conductor when the current flowing

through the other conductor is 5 KA.


F = 20.4 x 9.81= 200.124 N

d=0.05 mI1 = 5000 A

vv

--1 2

12 21 1 2

4 10 I I L LF =F = =2 10 I I

2 d d


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Chapter 3

LIFTING POWER OF MAGNET


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N

S

A = area of cross section ofeach pole

P = force in Newtonbetween two poles

N

S

P


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N

S

A = area of cross section ofeach pole

P = force in Newtonbetween two poles

N

S

P


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N

S

A = area of cross section of each poleP = force in Newton between two

poles

N

S

dxWork done= P . dx

Volume increased= A . dx

P


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Energy in a

Magnetic Field Let the energy stored in the inductor at

any time be E.

E = U = L I2


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Energy stored in

a Magnetic Field Given E = U = L I2,

So, the energy stored per volume

This applies to any region in which a magneticfield exists not just the solenoid

22

21

2 2o

o o

B BU n A A

nQ

Q Q

! !

l l

2

2B

o

U Bu

V Q! !


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N

S

dx

Energy stored into air magneticfield = B2/ 20 joule / m

3

So energy stored into airmagnetic field

= B2 {A . dx} / 20

Work done= B2 {A .dx} / 20 = P . dx

So, the lifting power of magnet= B2A/ 20

P


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Electromagnetism and magnetic circuit 6

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