wu4c10 by adel khamis

Unit Four -438- Chapter Ten

Magnetic Effect of an Electric Current

Christian Oersted:

In 1819, Oersted discovers that the electric current produce magnetic field.

The Magnetic flux intensity (B):

Magnetic field consists of lines called ………………………………

The total number of flux lines () has unit of …………

Magnetic flux density (B) is: …………………………………………….

Rule: ………………………

Its unit is …………………………. or ……………..……………….

Magnetic field due to a current in a straight wire:

Magnitude:

The relation between magnetic field intensity and the electric current intensity

is …………….…….

B ………

The relation between magnetic

field intensity and the distance is

…………….…….

B ………

Where:

: The permeability of the medium, and its value for air or space equals

4 x10-7 and its unit is ………………. or ………………….

I: means ………………………….

Work Sheet 2008/2009


d: means ……………………………………………………………….

Shape:

It is …………………………………………………………………..

The circular magnetic flux lines are ………. near the wire, and

…………. as the distance from the wire increases.

Direction:

It can be indicated by using of three methods:

a) Ampere’s right hand rule:

……………………………………………………

……………………………………………………

……………………………………………………

……………………………………………………

b) Using a small compass:

……………………………………………………

……………………………………………………

……………………………………………………

c) Right hand screw rule:

……………………………………………………

……………………………………………………

……………………………………………………

……………………………………………………



Neutral point:

Defination: …………………………………………………………………..

In case of two wires carrying current in the same direction the newtral point

will be (between them/ outside the wires)

In case of two wires carrying current in opposite direction the newtral point

will be (between them/ outside the wires)

Neutral point form close to the wire that carry (low, high) current.

Example:

1. Determine the magnetic flux density at a distance of 10 cm in air from the

center of a long wire carrying a current of 10 A. = 4 x 10-7 wb/A.m

…………………………………………………………………………………

…………………………………………………………………………………

…………………………………………………………………………………

…………………………………………………………………………………

[2 x 10-5 tesla]

2. Two straight line wires, the first carry 2A and the second carry 8A. Find

the distance between the two wires if the neutral point between them at 4

cm apart from the first wire.

…………………………………………………………………………………

…………………………………………………………………………………

…………………………………………………………………………………

…………………………………………………………………………………

[20 cm]



3. Two straight line wires, the first carrying 2 A and the distance between

them is 50 cm. if the neutral point was at 10 cm apart from the first wire

when the current in both of them was at the same direction find the

position of the neutral point if the current in one of them is inverted.

…………………………………………………………………………………

…………………………………………………………………………………

…………………………………………………………………………………

…………………………………………………………………………………

[16.6 cm]

Magnetic field due to current in a circular loop:

The Magnitude:

The relation

between magnetic field

intensity and the electric

current intensity is …………….…….

B ………

The relation between magnetic field intensity and the radius is …………….

…….

B ………

Where:

: …………………………………………………………………….

N: ……………………………………………………………………..

I: ……………………………………………………………………..



r: ……………………………………………………………………..

Shape:

………………………………………………………………………………..

………………………………………………………………………………..

Direction:

………………………………………………………………………………..

………………………………………………………………………………..

………………………………………………………………………………..

Example:

1. Determine the magnetic flux density at the center of a circular loop of

radius 11 cm carrying a current 1.4 A. If the wire consists of a coil having

20 turns.

………………………………………………………………………………..

………………………………………………………………………………..

………………………………………………………………………………..

[1.6 x 10-4 tesla]

2. circular coil consists of 20 loops carrying current of 1A, find the current

that pass in a straight line wirre tangent to the coil to vanish the flux

intensity at the center of the circular coil.

………………………………………………………………………………..

………………………………………………………………………………..

………………………………………………………………………………..

[62.8A]



Magnetic field due to current in a solenoid:

Magnitude:

The relation between magnetic field intensity and the electric current intensity is

…………….……., B ………

The relation between magnetic field intensity and the length of the solenoid is

…………….……., B ………

Where:

: …………………………………………………………………….

N: ……………………………………………………………………..

I: ……………………………………………………………………..

L: ……………………………………………………………………..

n: ……………………………………………………………………..



Shape:

…………………………………………………………………………………

……………………………………

Direction:

…………………………………………………………………………………

…………………………………….

Examples:

1. A long solenoid has 800 turns. The current flowing through the wire is

0.7 A. Find the magnetic flux density in the interior of the solenoid. Knowing that

its length is 20 cm.

…………………………………………………………………………………

…………………………………………………………………………………

…………………………………………………………………………………

[3.52 x 10-3 Tesla]

2. A solenoid is constructed by winding 800 turns of wire on a 20 cm iron

core. What current is required to produce a flux density of 0.815 Tesla in the

center of the solenoid? The permeability of iron is 1.63 x 10-2 wb/A.m

…………………………………………………………………………………



…………………………………………………………………………………

…………………………………………………………………………………

[1.25 x 10-2 A]

Force on a current carrying wire in a

magnetic field:

…………………………………………………………………………………

……………………………………………………………………

Direction of the motion:

The figure (A) shows the field

between the two poles of U shaped magnet.

The figure (B) shows the shaped of the

magnet field resultant from ……………………………..

When the wire is placed in the magnetic field of the magnet, the flux line

will be close to each other at the side at which ………………………….., and far

a part at a side at which …………………………………………...

Because the flux lines have ……………. force between each other, the

wire will move from the side at which …………………………. to the side at

which ……………………………………..

Direction of motion can be determined by using of Fleming’s left hand

rule:



Fleming’s left hand rule:

…………………………………………………………………………………

…………………………………………………………………………………………

……………………………………………



Factor affected on the magnetic motive force:

The length of the wire (L):

……………………………………………….

F ……..

The current (I):

…………………………………………………………….

F ……..

The magnetic flux density (B):

…………………………………………….

F ……..

Therefore:

F ……..

F = ……..

By using of ……….. as a unit of force, ……….. as a unit of magnetic flux

density, ………. as a unit of electric current intensity and ………. as a unit of length,

The constant is equal to one.

F = ……..

If the field is inclined by angle of to the direction of the wire

F = B I L sin

Define tesla: ……………………………………………………………..

The force between two wires carrying current in the same direction will be

(inward/ outward)

The force between two wires carrying current in opposite direction will be

(inward/ outward)



For two wires A and B the mutual force between them can be calculated from

the relation: F = B I L sin

where:

F is ………………………………. and its unit is ……………………………….

B is ………………………………. and its unit is ……………………………….

I is ………………………………. and its unit is ……………………………….

L is ………………………………. and its unit is ……………………………….

Examples:

1. Find the unit and the dimensional formula of the magnetic flux density.

…………………………………………………………………………………

…………………………………………………………………………………

2. A wire 30 cm long supports a current of 4 amperes in a direction

perpendicular to a magnetic field. If the force on the wire is 6 Newton find the

magnetic flux density.

…………………………………………………………………………………

…………………………………………………………………………………

…………………………………………………………………………………

…………………………………………………………………………………

[5 Tesla]

3. If the wire in the previous example makes an angle of 30 with respect to

the field, find the force acting on the wire.

…………………………………………………………………………………

…………………………………………………………………………………



…………………………………………………………………………………

[3 Newton]

4. Find the mutual force between two parallel wires; the first has length of

50 cm and carrying current of 2A while the second and length of 1 m and carrying

current of 8 A. Given that the distance between them is 20 cm and of air is

4x10-7.

…………………………………………………………………………………

…………………………………………………………………………………

…………………………………………………………………………………

…………………………………………………………………………………

[8x10-6 Newton]

Torque on a rectangular current carrying loop:

In case of rectangular current carrying loop, there is a

force per each side of the rectangular coil.

The force acts per each horizontal side will

……………. because ………………………

The double force act on the vertical sides cause a torque that can

calculated by:

= F1 d1 + F2 d2

Where:

d1 ……………………………………………………………………………..

d2 ……………………………………………………………………………..

F1 ……………………………………………………………………………..


d

L


F2 ……………………………………………………………………………..

d ……………………………………………………………………………..

But in this case the two forces are equal in magnitude, and the center of

rotation at the center of d

= …………………………..

but F = BIL

= ………………………….

but A = L d

= …………………………

In case of N turns

= BIAN

= BIAN cos

= BIAN Sin

Where:

: is the angle between the

magnetic field and the plane of the

coil.

is the angle between the

magnetic field and the normal to plane of the coil.

The direction of circular motion is normal to the coil plane and the

magnetic field is normal to the force therefore



the angle between the force and the direction of motion is equal to the angle

between the plane of the coil and the field.

Example:

Find the torque of coil has 200 turns and carry 0.5 A, placed in magnetic field

of 10 wb, then find the angle between the coil and the field at which the torque

decrease to half its value.

…………………………………………………………………………………………

…………………………………………………………………………………………

…………………………………………………………………………………………

…………………………………………………………………………………………

Electrical Measuring Instruments

Moving coil galvanometer (sensitive galvanometer):

Its operating principle is based upon:

…………………...

…………………………………………………………..

It used to measure …………………………….

It is also used to …………………………………….

Structure:

…………………………………………………..

…………………………………………………..

…………………………………………………..

…………………………………………………..

………………………………………………….

………………………………………………….



The springs are used to:

a) …………………………………..

b) …………………………………...

c) …………………………………...

G.R.:

The permanent magnet is concave.

a) …………………………………..

b) …………………………………...

The galvanometer sensitivity:

Definition: ………………………………………………………………

It depends on …………………… and …………………….

Its unit is: ………………………………………………………………

Example:

Sensitive galvanometer has a sensitivity of 50 A per scale division. What

current is required to give full deflection of the pointer through 25 scale divisions to

the right or left of the equilibrium position?

…………………………………………………………………………………………

…………………………………………………………………………………………

[1.25x10-3A]

Direct Current Ammeter:

It based on

…………………………………………………………………………



Ammeter is a galvanometer connected with …………. resistance in

…………….. called …………………. (R…….)

Ammeter is connected to the electric circuit in …………………...

I = …..……… + …………. (1)

V : …………. = ………… (2)

From ohm’s law:

V = ………………………. (3)

From (2) and (3):

………………. = ………………..

From (1)

Is = ………… - …………..

Benefits of the shunt:

1. …………………………………………………………………

2. …………………………………………………………………

3. …………………………………………………………………

Example:

A certain galvanometer has a resistance 2 , and a current of 5 m A is

required for full-scale deflection. What shunt resistance must be used to convert the

galvanometer to an ammeter whose maximum range is 10 amperes?

…………………………………………………………………………………………

…………………………………………………………………………………………


G

Shunt

Ig

IS

I


…………………………………………………………………………………………

…………………………………………………………………………………………

[10-3]

In the previous problem, calculate the maximum current can be measured if

the shunt is replaced by another one of resistance 0.1 x 10-3

…………………………………………………………………………………………

…………………………………………………………………………………………

…………………………………………………………………………………………

…………………………………………………………………………………………

N.B.:

By decrease the shunt, I max can be measured is …………………….

I max is (directly / inversely) proportional to the shunt resistance

Sensitivity of the ammeter is (directly / inversely) propoertioinal to the shunt

resistance.

Direct current voltmeters:

It is based on

…………………………………………………………………….

Voltmeter is a galvanometer connected with ………….. Resistance in

…………… called ……………………... (R…….)

Voltmeter is connected to the electric circuit in ……………………….


Gmultiplier

Ig

Im

Electric circuit


V = …………. + ………….. (1)

I : ……………. = …………… (2)

From (1)

Vm = ………………… (3)

From Ohm’s law:

Vm = ………………… (4)

From (3) and (4):

………………….. = ………………….

From (2):

Im = ………….

Benefits of the multiplier:

1. …………………………………………………………………

2. …………………………………………………………………

3. …………………………………………………………………

Example:

A galvanometer has an internal resistance of 0.1 and gives a full scale

deflection for a current of 1 m A. Calculate the multiplier resistance necessary to

convert this galvanometer to a voltmeter whose maximum range is 50 volts.

…………………………………………………………………………………………

…………………………………………………………………………………………



…………………………………………………………………………………………

…………………………………………………………………………………………

…………………………………………………………………………………………

[49999.9 ]

How to use the pervious voltmeter to measure 50 A.

…………………………………………………………………………………………

…………………………………………………………………………………………

…………………………………………………………………………………………

…………………………………………………………………………………………

…………………………………………………………………………………………

[1 ]

Questions:

Draw a graphical relation between I max and Ig of an ammeter, then mention the

slop indecation

………………………………………………………

………………………………………………………

Draw a graphical relation between V max and Ig of a voltmeter, then mention the

slop indecation

………………………………………………………

………………………………………………………

The Ohmmeter:

It is based upon: ………………..…………………

………………………………………………………..



It consists of:

……………………………………………………………………………

…………………………………………………

The battery used has electromotive force of 1.5 volt and the micro Ameter

obtain full scale deflection at 400 micro ampere therefor, to obtain full scale

deflection of the micro ammeter, it require total resistance of …………….. .

Since the internal resistance of the microameter is 250 and the fixed

resistance is 3000 there for the variable resistance must adjust to …… to

obtain full scale deflection when no external resistance connected.

By connecting of an external resistance the pointer will point law deflation and

the external resistance can be measured.

Questions:

what is the benefit of rheostate in the ohmmeter?

……………………………………………………………………………………

……………………………………………………………………………………..

G.R.:

The scale of the ohmmeter runs backward.

…………………………………………………………………………………………

The scale is non uniform.

…………………………………………………………………………………………



Example:

A Millie ammeter of resistance 5 its coil is capable to carry a current of 15 mA.

It is desired to use it as an ohmmeter using an electric cell of 1.5 V and internal

resistance of 1 . Calculate the required standard resistance and calculate the

external resistance when connected to it, the pointer indicates 10 mA? Calculate

the current that flows through it when connected to an external resistance of

400?

……………………………………………………………………………………

……………………………………………………………………………………

……………………………………………………………………………………

……………………………………………………………………………………

…………………………………………………..……………………………….

……………………………………………………………………………………

….

[94, 50, 0.003 A]

The following table represent the relation between the external resistance and the

current pass through the micro ammeter, draw the graphical relation between I at

y-axis and R at x-axis and from the graph find the internal resistance of the

ohmmeter and the electromotive force of its battery

Rex 0 120 300 410 500 690

I 0.01 0.008 0.00625 0.0055 0.005 0.0042

……………………………………………………………………………………

……………………………………………………………………………………

……………………………………………………………………………………

……………………………………………………………………………………

…………………………………………………..……………………………….



……………………………………………………………………………………

….


wu4c10 by adel khamis

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