Pdt Unit14 Students)

A group of phenomena associated with magnetic field.

UNIT 14: MAGNETISM

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Learning outcomes:

14.1

14.2

14.3

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Magnetic field




Explain magnetic field




Identify magnetic field sources




Sketch the magnetic field lines.




Determine the direction of magnetic field

Force on a moving charged particle in a uniform magnetic field

Force on a currentcarrying conductor in a uniform magnetic field

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Explain the effect on a charge moving in a uniform magnetic field

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Use F = qvB sin to explain the factors affecting the magnitude of the force



Explain the effect on a currentcarrying conductor in a uniform magnetic field



Determine the direction of the force using both Right Hand Rule and Fleming’s Left Hand Rule.



Use F = IlB sin to explain the factors affecting the magnitude of the force.

θ

θ

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14.1 Magnetic Field 

 



Definition –

the region around a magnet where a magnetic force can be experienced. A stationary electric charge is surrounded by an electric field only. When an electric charge moves, moves it is surrounded by an electric field and a magnetic field. The motion of the electric charge produces the magnetic field. field Magnetic field has two poles; north (N) and south (S). (S) The magnetic poles are always found in pairs. A single magnetic pole has never been found. Characteristics:  Like poles (N-N or S-S) repel each other.  Opposite poles (N-S) attract each other.

14.1.1 Magnetic field lines  

Magnetic field lines are used to represent a magnetic field. By convention, magnetic field lines leave the north pole and enters the south pole of a magnet.  Magnetic field lines can be represented by straight lines or curves. The tangent to a curved field line at a point indicates the direction of the magnetic field at that point as shown in figure 14 a. P

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Fig. 14 a

direction of magnetic field at point P.

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•

•

Magnetic field can be represented by crosses or by dotted circles as shown in figures 14 b and 14 c. X

X

X

X

X

X

X

X

X X X X Fig. 14 b : magnetic field lines enter Fig. 14 c : magnetic field lines leave the the page perpendicularly page perpendicularly A uniform field is represented by parallel lines of force. This means that the number of lines passing perpendicularly through unit area at all cross-sections in a magnetic field are the same as shown in figure 14 d. Fig. 14 d

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unit cross-sectional area A non-uniform field is represented by non-parallel lines. The number of magnetic field lines varies at different unit cross-sections as shown in figure 14 e. 4

Fig. 14 e

weaker field in A2 A1

A2

stronger field in A1 The number of lines per unit cross-sectional area is proportional to the magnitude of the magnetic field. field (N α B )  Magnetic field lines do not intersect one another. 14.1.2 Magnetic field lines Pattern  The pattern of the magnetic field lines can be determined by using two methods.  Using compass needles (shown in figure 14 f) 

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Fig. 14 f : plotting a magnetic field line of a bar magnet.

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Using sprinkling iron filings on paper (shown in figure 14 g).



Fig. 14 g : Thin iron filing indicate the magnetic field lines around a bar magnet. Figure below shows the various pattern of magnetic field lines around the magnets.

a. Bar magnet

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b. Horseshoe or U magnet

c. Two bar magnets (unlike pole) - attractive

d. Two bar magnets (like poles) - repulsive

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Neutral point (point where the resultant magnetic force is zero)

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14.1.3 Direction of magnetic field. Magnetization of a Soft Iron  There are two methods to magnetized the soft iron.  Using the permanent magnet.  One permanent magnet  A permanent magnet is bring near to the soft iron and touching the surface of the soft iron by following the path in the figure 14 h.

N

 

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S

Fig. 14 h This method is called induced magnetization. The arrows in the soft iron represent the magnetization direction with the arrowhead being the north pole and arrow tail being the south pole. It is also known as domains ( the tiny magnetized region because of spin magnetic moment of the electron). 8

In an demagnetized piece of soft iron, these domains are arranged randomly but it is aligned in one direction when the soft iron becomes magnetized.  The soft iron becomes a temporary magnet with its south pole facing the north pole of the permanent magnet and vise versa as shown in figure 14h. Two permanent magnets  Bring and touch the first magnet to one end of the soft iron and another end with the second magnet as shown in figure 14i and 14j. 



Fig. 14 i

N

S SF027

Fig. 14 j

S

N

N

S 9



Using Electrical circuit.  A soft iron is placed inside a solenoid (a long coil of wire consisting of many loops of wire) that is connected to the power supply as shown in figure 14k. N

SS

I

I Current anticlockwise





Switch, S

Current - clockwise

Fig. 14 k When the switch S is closed, the current I flows in the solenoid and produces magnetic field. The directions of the fields associated with the solenoid can be found by viewing the current flows in the solenoid from both end as shown if figure 14k or applying the right hand grip rule below. Important Thumb – north pole

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Other fingers – direction of current in solenoid.

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Other examples: a.

S

N

I

b.

I

S

I



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N

I

Note :  If you drop a permanent magnet on the floor or strike it with a hammer, you may jar the domains into randomness. randomness The magnet can thus lose some or all of its magnetism.  Heating a magnet too can cause a loss of magnetism.

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The permanent magnet also can be demagnetized by placing it inside a solenoid that connected to an alternating source. source Example 1 : (exercise) Sketch the magnetic field lines pattern around the bar magnets as shown in figures below. a. b.

 

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When a current flows in a conductor wire or coil, the magnetic field will be produced.  The direction of magnetic field around the wire or coil can be determined by using the right hand grip rule as shown in figure 14k. Magnetic field of a long straight conductor (wire) carrying current  The magnetic field lines pattern around a straight conductor carrying current is shown in figures 14l and 14m. 

I

r B

r B

I

or

r B

r B

I

Fig. 14 l Current out of the page

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r B

I r B

r B

I

or

r B

Fig. 14 m X

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XI

Current into the page

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Magnetic field of a Circular Shaped Coil  The magnetic field lines pattern around a circular shaped coil carrying current is shown in figures 14n.

I R N

N or

S

I

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I

X

I

S

I

Fig. 14 n

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Magnetic field of a Solenoid  The magnetic field lines pattern around a solenoid carrying current is shown in figures 14o. N

S

I

I or Fig. 14 o XI

XI

XI

XI

N

S

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I

I

I 16



The equation of magnetic flux density around a solenoid carrying current I can be determined by using Ampere’s law.  Consider a very long solenoid with closely packed coils, the field is nearly uniform and parallel to the solenoid axes within the entire cross section, as shown in figure 14p. Fig. 14 p

r r d c B dl r r r dl dl dl a b r l B(outside) = 0(very small ) 

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To find the magnetic field inside the solenoid and at the centre, we choose and draw the rectangle closed path abcd as shown in figure 14p (clockwise) for applying Ampere’s law. By considering this path consists of four segment : ab, bc, cd and da, then Ampere’s law becomes

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Example 2 : Two long straight wires are placed parallel to each other and carrying the same current, I. Sketch the magnetic field lines pattern around both wires a. when the currents are in the same direction. b. when the currents are in opposite direction. Solution: a. I

I

I

I

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I

or

I

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b.

I I

I I or

I

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XI

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14.2 Magnetic Force on a Moving Charge 



A stationary electric charge in a magnetic field will not experience any force. force But if the charge is moving with a velocity, v in a magnetic field, B then it will experience a force. This force known as MAGNETIC FORCE. FORCE In vector form, form (14.2 a)



The magnitude of the magnetic force can be calculated by using the equation below

(14.2 b)

where

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F : magnetic force B : magnetic flux density v : velocity of a charge q : magnitude of the charge r r θ : angle between v and B 20



The direction of the magnetic force can be determined by using the Fleming’s hand rule.  Fleming’s right hand rule : - for negative charge shown in figures 14q and 14r  Fleming’s left hand rule : - for positive charge

Fig. 14 q Thumb –

Important

Fig. 14 r

First finger – 

Second finger – Example 3 : r Determine the direction of the magnetic force, F exerted on a charge in each problems below. r a. b. v r

+

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r B

B −

r v

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c. X

X

X

X

rX

X

X

X

d.

− X

X X

I

Solution: a. By using Fleming’s left hand rule,

b. By using Fleming’s right hand rule,

r B −

r B c. By using Fleming’s right hand rule, X

X

X

r v

d. Using right hand grip rule to determine the direction of magnetic field forms by the current I on the charge position. Then apply the Fleming’s right hand rule,

− X

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r BX

X X

rX v

X

X

X

r v

I

r v

+

X

−r v

+

v

r BX

e.

X X

X

−r I v

X

X

X

X

r B 22

e.

Using right hand grip rule to determine the direction of magnetic field forms by the current I on the charge position. Then apply the Fleming’s left hand rule, thus

r X XB X Xr v + I 

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X

X

X

Example 4 : (exercise) Determine the sign of a charge in each problems below. a. b. r r

F



X

B r v

r F

r B

r v

Ans. : positive charge, positive charge Example 5 : (exercise) Determine the direction of the magnetic force exerted on a positive charge in each figures below when a switch S is closed. 23

a.

b.

+ Switch, S



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r v

r v

+

Switch, S

Ans. :a. into the page, b. out of page Example 6 : Calculate the magnitude of the force on a proton travelling 3.0 x107 m s-1 in the uniform magnetic flux density of 1.5 Wb m-2, if : a. the velocity of the proton is perpendicular to the magnetic field. b. the velocity of the proton makes an angle 50° with the magnetic field. (Given the charge of the proton is +1.60 x 10-19 C) Solution: v=3x107 m s-1, B=1.5 T, q=1.60x10-19 C a. Given θ = 90° then by applying the equation of magnetic force on a moving charge, thus ∴ F = 7.2 x10 −12 N

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b. Given θ = 50° then by applying the equation of magnetic force on a moving charge, thus

F = Bqv sin θ F = 5.5 x10 −12 N





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Example 7 : (exercise) An electron experiences the greatest force as it travels 2.9 x 106 m s-1 in a magnetic field when it is moving north. The force is upward and of magnitude 7.2 x 10-13 N. Find the magnitude and direction of the magnetic field. (Giancolli, pg.705, no.22) (Given the charge of the electron is -1.60 x 10-19 C) Ans : 1.6 T to the east. Example 8 : (exercise) An electron is moving in a magnetic field. At a particular instant, the speed of the electron is 3.0 x 106 m s-1. The magnitude of the magnetic force on the electron is 5.0 x 10-13 N and the angle between the velocity of the electron and the magnetic force is 30°. Calculate the magnitude of the magnetic field . (Given the charge of the electron is -1.60 x 10-19 C) Ans : 1.2 T

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14.3 Magnetic Force on a current-carrying conductor  



When a current-carrying conductor is placed in a magnetic field B, thus a magnetic force will acts on that conductor. In vector form, form (14.3 a) The magnitude of the magnetic force exerts on the current-carrying conductor is given by (14.3 b) where

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F : magnetic force B : magnitude of the magnetic flux density I : current L : length of the conductor r θ : angle between direction of I and B

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The direction of the magnetic force can be determined by using the Fleming’s left hand rule as shown in Figure 14 s

Fig. 14 s

Important Thumb – First finger – Second finger –

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Note : 

From eq. (14.3 a),  the magnetic force on the conductor has its maximum value when the conductor (and therefore the current) and the magnetic field are perpendicular (at right angles) to each other then θ=90° (shown in figure 14 t).  the magnetic force on the conductor is zero when the conductor (and therefore the current) is parallel to the magnetic field then θ=0° (shown in figure 14 u).

r B

θ = 90 o I Fmax = BIL sin 90 o Fmax = BIL Fig. 14 t 

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r B

I

θ = 0o

F = BIL sin 0 o

F =0

Fig. 14 u

One tesla is defined as the magnetic flux density of a field in which a force of 1 newton acts on a 1 metre length of a conductor which carrying a current of 1 ampere and is perpendicular to the field. 28



Example 15 : Determine the direction of the magnetic force, exerted on a conductor carrying current, I in each problems below. a. b. X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

r BX

r BX

I X

X

I X

X

X

Solution: For both problems, use Fleming’s left hand rule : a. b. X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

r BX

r BX

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I X

X

I X

X

X

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Example 16 : A wire of 20 cm long is placed perpendicular to the magnetic field of 0.40 Wb m-2. a. Calculate the magnitude of the force on the wire when a current 12 A is flowing. b. For the same current in (a), determine the magnitude of the force on the wire when its length is extended to 30 cm. c. If the force on the 20 cm wire above is 60 x 10-2 N and the current flows is 12 A, find the magnitude of magnetic field was supplied. Solution: L=20x10-2 m, B=0.40 T, θ=90° a. Given I = 12 A. By applying the equation of magnetic force on a current-carrying conductor, thus

F = BIL sin θ

b. Given I = 12 A and L = 30x10-2 m By applying the equation of magnetic force on a current-carrying conductor, thus

F = BIL sin θ SF027

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c. Given I = 12 A, L = 20x10-2 m , θ=90° and F = 60x10-2 N By applying the equation of magnetic force on a current-carrying conductor, thus F = BIL sin θ

F B= IL sin θ THE END

6.8 Forces between two current-carrying conductors 

Consider two identical straight conductors X and Y carrying currents I1 and I2 with length L are placed parallel to each other as shown in figure 14v.

X

Y

I1

r B2 SF027 Fig. 14 v

r F12

I2 r B1 P

 

µ0 I 1 Direction : into the B1 = 2πd page/paper

r F21

Q



I1

d

I2

The conductors are in vacuum and their separation is d. The magnitude of the magnetic flux density, B1 at point P (on conductor Y) due to the current in conductor X is given by

Conductor Y carries a current I2 and in the magnetic field B1 then conductor Y will experiences a magnetic force, F12.

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