PHYSICS
CHAPTER 6
CHAPTER 6: Magnetic field (7 Hours)
PHYSICS CHAPTER 6 Learning Outcome: s c i s y h p / y m . .. u d e . . ki r t a m . . h p m . .k w w w
6.1 6. 1 Ma Magn gnet etic ic fi fiel eld d (1 hour) At the end of this cchapter, hapter, students should be able to: this chapter, Define magnetic field.
Identify magnetic field sources. Sketch the magnetic field lines.
PHYSICS
CHAPTER 6
6.1 Magnetic field
is defined as a region around a magnet where a magnetic force can be experienced. experienced . A stationary electric charge is surrounded by an electric field only. only. When an electric charge moves, moves , it is surrounded by an electric field and a magnetic field. field . The motion of the electric charge produces the magnetic field. field . Magnetic field has two poles, called north (N) and south (S). (S). This magnetic poles are always found in pairs whereas a single magnetic pole has never been found. Like poles (N-N or S-S) repel each other. Opposite poles (N-S) attract each other.
PHYSICS
CHAPTER 6
6.1 .1.1 .1 Magn gne eti tic c fi fie eld li line nes s
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 6.1. direction of magnetic field at point P.
P Figure 6.1
Magnetic field can be represented by crosses or by dotted circles as shown in Figures 6.2a and 6.2b.
X
X
X
X
X
X
X
X
Figure 6.2a : magnetic field lines enter the page perpendicularly
Figure 6.2b : magnetic field lines leave the page perpendicularly
PHYSICS
CHAPTER 6
A uniform field is represented by parallel lines of force. 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 6.3.
unit cross-sectional area Figure 6.3
A non-uniform field is represented by by non-parallel lines. The number of magnetic field lines varies at different unit crosssections as shown in Figure 6.4.
PHYSICS
CHAPTER 6
weaker field in A2
A1
A2
stronger field in A1 Figure 6.4
The number of lines per unit cross-sectional area is proportional to the magnitude of the magnetic field . Magnetic field lines do not intersect one another.
PHYSICS
CHAPTER 6
6.1. 6. 1.2 2 Ma Magn gnet etic ic fi fiel eld d lin lines es pa patt tter ern n
The pattern of the magnetic field lines can be determined by using two methods. compass needles (shown in Figure 6.5)
Figure 6.5: plotting a magnetic field line of a bar bar magnetic. sprinkling iron filings on paper (shown paper (shown in Figure 6.6).
Figure 6.6: thin iron filing indicate the magnetic field field lines.
PHYSICS
CHAPTER 6
Figures 6.7 shows the various pattern of magnetic field lines around the magnets. a. Bar magnet
Figure 6.7a b. Horseshoe or U magnet
PHYSICS
CHAPTER 6
c. Two bar magnets ( unlike pole) pole) attractive
Figure 6.7c d. Two bar magnets ( like poles) poles) - repulsive
Neutral point (point where Figure 6.7d the resultant magnetic force is zero). zero ).
PHYSICS
CHAPTER 6
6.1 .1.3 .3 Eart rth h’s mag magn neti tic c fie field
The Earth’s magnetic field is like that of a giant bar magnet as illustrated in Figure 6.8 with a pole near each geographic pole of the Earth. The magnetic poles are tilted away from the rotational axis by an angle of 11.5°.
Since the north pole of a compass needle (Figure 6.8) points toward the north magnetic pole of the Earth, and since opposite attract, it follows that
the north geographical pole of the Earth is actually near the south pole of the Earth’s magnetic field. field .
Figure 6.8
Figure 6.8 also shows that the field lines are essentially horizontal (parallel to the Earth’s surface) near the equator but equator but enter or leave the Earth vertically near the poles
PHYSICS
CHAPTER 6
6.1. 6. 1.4 4
Magn Ma gnet etiz izat atio ion n of of a So Soft ft Ir Iron on
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 6.9.
Figure 6.9
N
S
This method is called induced magnetization. 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
PHYSICS
CHAPTER 6
In an unmagnetized 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 vice versa as shown in Figure 6.9. 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 6.10.
S
NN Figure 6.10
S
PHYSICS
CHAPTER 6
Using the 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 supply as shown in Figure 6.11.
N
S I
I Current anticlockwise
Switch, S
Current - clockwise
Figure 6.11 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 or applying the right hand grip rule as shown in Figure
PHYSICS
CHAPTER 6
Other examples:
S I Note: Thumb – north pole
N S I
N I
I
Other fingers – direction of current Figure 6.12a Figure 6.12b in solenoid. solenoid. 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. The permanent magnet also can be demagnetized by placing it inside a solenoid that connected to an alternating source .
PHYSICS 6.1 .1.5 .5
CHAPTER 6 Mag agne neti tic c flu flux x den dens sit ity y, B
is defined as the magnetic flux per unit area across an area at right angles to the magnetic field. field . Mathematically,
B =
Φ
(6.1)
A⊥
where
Φ : magnetic flux A⊥ : area at right angles to the the magnetic field
It also known as magnetic induction (magnetic field intensity OR strength) strength) It is a vector quantity and its direction follows the direction of the magnetic field. field . Its unit is tesla (T) OR weber per metre squared (Wb m 2). Unit conversion : 4 −2
1 T = 1 Wb m
= 10
gauss(G)
PHYSICS
CHAPTER 6
Example 1 : The direction of any magnetic field is taken to be in the direction that an Earth-calibrated compass points. Explain why this mean that magnetic field lines must leave from the north pole of a permanent bar magnet and enter its south pole. Solution :
Near the north pole of a permanent bar magnet, the north pole of a compass will point away from the bar magnet so the field lines leave the north pole. Near the south pole of a permanent bar magnet, the north pole of a compass will point toward the bar magnet so s o the field lines enter the south pole.
PHYSICS
CHAPTER 6
Exercise 6.1 : 1.
Sket Sketch ch the the magn magnet etic ic fiel field d lin lines es patt pattern ern aroun around d the the bar bar magnets for following cases. a.
b.
PHYSICS CHAPTER 6 Learning Outcome: s c i s y h p / y m . .. u d e . . ki r t a m . . h p m . .k w w w
6.2 Magnet Magnetic ic produ produced ced by curr current ent-ca -carry rryin ing g conductor (1 hour) At the end of this cchapter, hapter, students should be able to: this chapter, Apply magnetic field formula
for a long straight wire,
B
=
2πr
for a circular coil,
B =
μ0 I
μ0 I 2 R
for a solenoid.
B
= μ0 nI
PHYSICS
CHAPTER 6
6.2 Magnetic field produced by current – carrying conductor
When a current flows in a conductor wire or coil coil,, the magnetic field will be produced. 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 6.13.
Note: Thumb – direction of current Other fingers – direction of magnetic field (clockwise OR anticlockwise)) anticlockwise Figure 6.13
PHYSICS
CHAPTER 6
6.2.1 6.2. 1 Ma Magn gnet etic ic fie field ld of of a long long str strai aigh ghtt cond conduc ucto tor r (wire) carrying current
The magnetic field lines pattern around a straight conductor carrying current is shown in Figures 6.14 and 6.15.
I
B
I
OR
B
B
I Figure 6.14
B
Current out of the page
PHYSICS
CHAPTER 6
I
B
B
B
X I
OR
I Figure 6.15
B
X Current into the page
Consider a straight conductor (wire) carrying a current I is placed in vacuum as shown in Figure 6.16.
PHYSICS
CHAPTER 6
r
Figure 6.16
I
P
X B
into the page (paper)
The magnitude of magnetic flux density (magnetic field intensity), B at point P at distance r from the wire carrying current is given by
B
=
μ0 I
2πr
(6.2)
−7 −1 μ : permeabili ty of free space = π × 4 10 T m A where 0 r : distance of a point from a straight conductor (wire)
PHYSICS 6.2. 6. 2.2 2
CHAPTER 6 Magn Ma gnet etic ic fie field ld of a cir circu cula larr coi coill
The magnetic field lines pattern around a circular coil carrying current is shown in Figures 6.17.
I
N
I OR
S I Figure 6.17
I
N X I
S
PHYSICS
CHAPTER 6
Consider a circular shaped conductor with radius R that carries a current I as shown in Figure 6.18.
The magnitude of magnetic field intensity B at point O ( centre of the circular coil or loop) loop ) , is given by
R O
B
=
μ0 NI NI 2 R
(6.3)
permea eabi bili lity of free space where μ0 : perm R : radius of the the circular coil N : number of coils (loops) Figure 6.18
I : current
PHYSICS 6.2. 6. 2.3 3
CHAPTER 6 Magn Ma gnet etic ic fi fiel eld d of of a so sole leno noid id
A solenoid is an electrical device in which a long wire has been wound into a succession of closely spaced loops with geometry of a helix. helix . The magnetic field lines pattern around a solenoid carrying current is shown in Figure 6.19.
N
S I
I Figure 6.19
PHYSICS
CHAPTER 6 OR
X I
X I
X I
X I
N
S I
I
I
I
PHYSICS
CHAPTER 6
The magnitude of magnetic field intensity at the centre (mid-point/ inside) of N N turn solenoid is given by
B = and
μ0 NI l N l
B
= μ0 nI
(6.4)
=n
The magnitude of magnetic field intensity at the end of N turn solenoid is given by
B
=
1 2
μ0 nI
(6.5)
where n : number of turns per unit length
PHYSICS
CHAPTER 6
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. directi on. Solution : a.
I
I
I
I
PHYSICS
CHAPTER 6
Solution : a.
OR
I
I
PHYSICS
CHAPTER 6
Solution : b.
I I
I I OR
I
X I
PHYSICS
CHAPTER 6
Example 3 : A long wire (X) carrying a current of 50 A is placed parallel to and 5.0 cm away from a similar wire (Y) carrying a current of 10 A. a. Determine the magnitude and direction of the magnetic flux density at a point midway between the wires : i. when the current are in the same direction. ii. when they are in opposite direction. b. When the currents are in the same direction there is a point somewhere between X and Y at which the magnetic flux density is zero. How far from X is this point ? (Given µ 0 = 4π × 10−7 H m−1)
PHYSICS
CHAPTER 6
Solution : I X
= 50 A; d = 5.0 × 10 −2 m; I Y = 10 A
a. i.
r X
d
BX
BX
A
r Y
OR
I X
A
r X
r Y
I Y
BY
BY I X
I Y
r X
d
= r Y = = 2.5 × 10− 2 m 2
By using the equation of magnetic field at any point near the straight wire, then at point A
BX : Magnitude of B
BX
=
μ0 I X 2πr X
( 4π × 10 − )50 = 2π ( 2.5 × 10 − ) − 7
BX BX
2
= 4.0 × 10 4 T
Direction : into the page OR upwards
PHYSICS
CHAPTER 6
Solution : I X
= 50 A; d = 5.0 × 10 −2 m; I Y = 10 A
BY : a. i. Magnitude of B
BY
=
μ0 I Y 2πr Y
( 4π × 10 − )10 = 2π ( 2.5 × 10 − ) − 7
BY BY
2
= 8.0 × 10 5 T
Direction : out of page OR downwards Therefore the totalmagnetic flux density at point A is
= BX + BY BA = − BX + BY BA
Note: Sign convention of B B: Out of page ⇒ positive (+) Into the page ⇒ negative ( −)
= −4.0 × 10 −4 + 8.0 × 10 −5 BA = −3.2 × 10 −4 T BA
Direction : into the page OR upwards
PHYSICS
CHAPTER 6
Solution : I X a. ii.
r X
= 50 A; d = 5.0 × 10 −2 m; I Y = 10 A d
BX BY A r
BY OR
Y
I X
I X
r X
BX
A
r Y
X
I Y
I Y
By using the equation of magnetic field at any point near the straight wire, then at point A
BX : Magnitude of B
( 4π × 10 − )50 = 2π ( 2.5 × 10 − ) 7
BX
2
BX
= 4.0 × 10−4 T
Direction : into the page OR upwards
PHYSICS
CHAPTER 6
Solution : I X
= 50 A; d = 5.0 × 10 −2 m; I Y = 10 A
BY : a. ii. Magnitude of B
( 4π × 10 − )10 = 2π ( 2.5 × 10 − ) − 7
BY BY
2
= 8.0 × 10 5 T
Direction : into the page OR upwards Therefore the magnetic flux density at point A is resultant
= BX + BY BA = − BX − BY
BA
= −4.0 × 10 −4 − 8.0 × 10 −5 BA = −4.8 × 10 −4 T BA
Direction : into the page OR upwards
PHYSICS
CHAPTER 6
Solution : I X b.
r X
= 50 A; d = 5.0 × 10 −2 m; I Y = 10 A
BX
BX C
d
r Y
OR
BY I X
I X
C
r X
= r r Y = d − r
r Y
I Y
BY
r X
I Y
Since the resultant magnetic flux density at point C is zero thus
= BX + BY 0 = − BX + BY μ0 I X BX = BY where BX = BC
2πr X
and
BY
=
μ0 I Y 2πr Y
PHYSICS
CHAPTER 6
Solution : I X b.
= 50 A; d = 5.0 × 10 −2 m; I Y = 10 A
μ0 I X 2πr X I X r 50 r
=
μ0 I Y
=
I Y
=
2πr Y
( d − r ) 10
(5.0 × 10 − − r ) 2
r = 4.2 × 10 −2 m
PHYSICS
CHAPTER 6
Example 4 : Two long straight wires are oriented perpendicular to the page as shown in Figure 6.20.
The current in one wire is I 1Figure = 3.0 A 6.20 pointing into the page and the current in the other wire is I 2= 4.0 A pointing out of page. Determine the magnitude and direction of the nett magnetic field intensity at point P. (Given µ 0 = 4π × 10−7 H m−1)
PHYSICS
CHAPTER 6
= 3.0 A; I 2 = 4.0 A; r 1 = 5.0 × 10 −2 m
Solution : I 1
= (5.0 × 10 ) + ( 5.0 × 10 ) B1 r = 7.1× 10−2 m 2 −2 r 5 . 0 10 r 1 θ × r 2 cos cos θ = 1 = r 2 7.1× 10− 2 cos θ = 0.704 704 I 1 X 5.0 10−2 m I 2 cos −2 × 5.0 × 10 sin sin θ = 704 − 2 = 0.704 7.1× 10 P
θ
B2
−2
r 2
2
−2
2
By applying the equation of magnetic field intensity for straight wire, −7 thus
B1 =
μ0 I 1
2πr 1
B1
( 4π × 10 )3.0 = 2π (5.0 × 10 − )
B1
= 1.20 × 10−5 T
2
PHYSICS
CHAPTER 6
= 3.0 A; I 2 = 4.0 A; r 1 = 5.0 × 10 −2 m ( μ0 I 2 4π × 10 −7 ) 4.0 B2 = B2 = 2πr 2 2π ( 7.1 × 10 − 2 ) −5 B2 = 1.13 × 10 T
Solution : I 1 and
Vector
B1
B2 Vector sum
x-component (T)
B1
= 1.20 × 10−5
y-component (T)
0
sin θ − B2 cos − B2 sin cos θ 704) 704) = −(1.13×10−5 )( 0.704 = −(1.13× 10−5 )( 0.704 = −7.96 × 10−6 = −7.96 × 10 −6 −6 −5 −6 B x = 1.20 × 10 − 7.96 × 10 B y = 0 − 7.96 × 10
= 4.04 × 10−6
= −7.96 × 10−6
PHYSICS
CHAPTER 6
Solution : I 1
= 3.0 A; I 2 = 4.0 A; r 1 = 5.0 × 10 −2 m
Therefore the magnitude of the nett magnetic field intensity at point P is given by
B = B x
2
+ B y 2
= ( 4.04 × 10 ) + ( − 7.96 × 10 ) −6 B = 8.93 × 10 T −6
2
−6
2
and its direction is
θ = tan tan B x −6 7 . 96 10 − × = tan −1 − 6 4.04 × 10
B y −1
B1
P
B2
θ = −63.1
(297 from +x-axis anticlockwise) OR
63.1
B
PHYSICS
CHAPTER 6
Example 5 : a. A closely wound circular coil of diameter 10 cm has 500 turns and carries a current of 2.5 A. Determine the magnitude of the magnetic field at the centre of the coil. b. A solenoid of length 1.5 m and 2.6 cm in diameter carries a current of 18 A. The magnetic field inside the solenoid is 2.3 mT. Calculate the length of the wire w ire forming the solenoid. (Given µ 0 = 4π × 10−7 T m A−1) Solution :
a. Given R =
10 × 10 −2 2
= 5.0 × 10 −2 m; N = 500; I = 2.5 A
By applying the equation for magnitude of the magnetic field at the centre of the circular coil, thus −7
B =
μ0 NI 2 R
( 4π × 10 )( 500 ) 2.5 B = 2( 5.0 × 10 − ) 2
−2
B 1 57 10
T
PHYSICS
CHAPTER 6
Solution : b. Given l = 1.5 m; r =
I = 18 A
2.6 × 10 −2 2
= 1.3 × 10 −2 m; Bi = 2.3 × 10 −3 T
By applying the equation of magnetic flux density inside the solenoid, thus −7 μ0 NI 4 10 π N (18) 3 − B i
=
l
2.3 × 10
( × =
)
1.5 N = 153 turns
Since the shaped for each coil in the solenoid is circle, then the
=
circumference 2πr circumference for one turn is
circumference = 2π (1.3 × 10 −2 ) circumference = 8.17 × 10 −2 m
= × L = 153 × (8.17 × 10 −2 ) L = 12 5 m
Therefore L the length the wire forming N (of circumfere nce ) the solenoid is
PHYSICS
CHAPTER 6
Exercise 6.2 : Given µ 0 = 4π × 10−7 T m A−1 1.
10.0 cm P1
5.0 cm
P2
15.0 cm
5.00 A
Figure 6.21
5.00 A
The two wires shown in Figure Figur e 6.21 carry currents of 5.00 A in opposite directions and are separated by 10.0 cm. a. Sketch the magnetic field lines pattern around both wires. b. Determine the nett magnetic flux density at points P1 and P2. ANS. : 1.33 10 5 T, out of page; 2.67 10 6 T, out of page
PHYSICS
CHAPTER 6
Exercise 6.2 : X
2.
X
Figure 6.22 Four long, parallel power wires each carry 100 A current. A cross sectional diagram for this wires is a square, 20.0 cm on each side as shown in Figure 6.22. a. Sketch the magnetic field lines pattern on the diagram. b. Determine the magnetic flux density at the centre of the square. ANS. : 4.0
10 4 T , to the left (180 )
PHYSICS CHAPTER 6 Learning Outcome: s c i s y h p / y m . .. u d e . . ki r t a m . . h p m . .k w w w
6.3 Force on a movi moving ng charge charged d particl particle e in a uniform uniform magnetic field (1 hour) At the end of this cchapter, hapter, students should be able to: this chapter, Use formulae: F = q v × B
(
)
Describe circular motion of a charge in a uniform magnetic field.
Use relationship F B = F C.
PHYSICS
CHAPTER 6
6.3 Force on a moving charged particle in a uniform magnetic field 6.3.1 Magnetic fo force
A stationary electric charge in a magnetic field will not experience a magnetic force. force . But if the charge is moving with a velocity, v in a magnetic field, B then it will experience a magnetic force. force. The magnitude of the magnetic force can be calculated by using the following equation:
F = qvB sin θ where F : magnetic force
B : magnetic flux density v : velocity of a charge q : magnitude of the charge θ : angle betw betwee een n v and and B
(6.6)
PHYSICS
CHAPTER 6
In vector form, form,
F = q (v × B )
(6.7)
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 6.23 Fleming’s left hand rule for positive charge and 6.24
B
F
F
v
v
Figure 6.23 Thumb – direction of Force First finger – – direction of Field Second finger – – direction of Velocity of Velocity
B
Note:
Figure 6.24
PHYSICS
CHAPTER 6
Example 6 : Determine the direction of the magnetic force, F exerted on a charge in the following problems: a.
b.
v
+
X X
B X
X X
−
X X
X
X
X
X
−
d.
X
v
v X
v
−
B c.
B
I e.
+ I
v
PHYSICS
CHAPTER 6
Solution : a. By using Fleming’s left hand rule, thus
F (into the page)
v
+
B
b. By using Fleming’s right hand rule, ru le, thus
B
v
−
(to the left) F
c. By using Fleming’s right hand rule, thus X
X
X
X
(to the left) F
B X
−
X
X
X
X
X
X
X
v X
PHYSICS
CHAPTER 6
Solution : d. Using right hand grip rule to determine the direction of magnetic field produces by the current I on the charge position. Then apply the Fleming’s right hand rule, thus
(to the left) F
X
X
X
X
X
I v
X
X
−
B
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
F (upwards) X X B X
X
X
X
+ I
v
PHYSICS
CHAPTER 6
Example 7 : Calculate the magnitude of the force on a proton travelling 5.0× 107 m s−1 in the uniform magnetic flux density of 1.5 Wb m −2, if : a. the velocity of the proton is i s 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 × 10−19 C) −1 −2 7 Solution : v 5.0 10 m s ; B 1.5 Wb m
= × a. Given θ = 90 Therefore F = qvB sin θ
=
= (1.60 ×10 ) ( 5.0 ×10 ) (1.5) sin 90 F = 1.20 ×10 −11 N −19
b. Given θ = 50 Hence
7
F = (1.60 ×10 ) ( 5.0 ×10 ) (1.5) sin 50 F = 9.19 ×10 −12 N −19
7
PHYSICS
CHAPTER 6
6.3.2 6.3. 2 Mo Motio tion n of a cha charg rged ed pa part rtic icle le in in a uni unifor form m magnetic field
Consider a charged particle moving in a uniform magnetic field with its velocity perpendicular to the magnetic field . As the particle enters the region, it will experiences a magnetic force which the force is perpendicular to the velocity of the particle. Hence the direction of its velocity changes but the magnetic force remains perpendicular to the velocity. v elocity. This magnetic force, F B makes the path of the particle is a circular as circular as shown in Figures 6.25a, 6.25b, 6.25c and 6.25d.
X
X
X
X
v
X
X F B
X
X
+
v
X
F B
X X
+
X X
X
+
X
v
+
v
v
+
F B
F B
+
v X
PHYSICS
v
−
CHAPTER 6 X
X
−
X
X
X
v
X
F B X
X F
B
X
X
v
X
−X
X
v
X
X
X
−
X
F B
v
−
Figure 6.25c
F B
−
v
Figure 6.25d
Since the path is a circle cir cle therefore the magnetic force F B contributes the centripetal force F c (nett force) in this motion. Thus
F B
= F c
qvB sin θ =
mv 2 r
and
θ = 90
PHYSICS
CHAPTER 6 r =
mv
(6.8)
Bq
the charged parti particl cle e where m : mass of the
v : magnitude of the velocity r : radius of the circular path q : magnitude of the charged particle The period of the circular motion, T makes by the particle is given by
v = r ω
v=
and
2π r T
ω =
2π T
T =
2π r
T =
2π m
v
and
r =
Bq (6.9)
Bq
And the frequency frequency of the circular circular motion is
mv
f =
1 T
PHYSICS
CHAPTER 6
Example 8 : v A
−
20.0 cm
B
Figure 6.26 An electron at point A in Figure 6.26 has a speed v of 2.50 × 106 m s-1. Determine a. the magnitude and direction of the magnetic field that will cause the electron to follow the semicircular path from A to B. b. the time required for the electron e lectron to move from A to B. (Given e=1.60× 10−19 C and me= 9.11× 10−31 kg)
PHYSICS
CHAPTER 6
Solution : v
= 2.50 × 10 6 m s −1; d = 20.0 × 10 −2 m
a. Since the path makes by the electron is a semicircular thus the the magnitude of the magnetic field is given by
r =
mv
Be d mv 2
=
and
r =
d
2 20.0 × 10 − 2 2
Be
( 9.11 × 10 − )( 2.50 × 10 ) = B (1.60 × 10 − ) 31
19
−4
B = 1.42 × 10 Direction of magnetic field : into the page OR
v A
−
B
F
6
B
T
PHYSICS
CHAPTER 6
Solution : v
= 2.50 × 10 6 m s −1; d = 20.0 × 10 −2 m
b. The period of the electron is
v = rω and ω = d 2π v = 2 T
2π T
2.50 × 10 6
( 20.0 × 10 − )π = 2
T −7 T = 2.51× 10 s
Since the path is the semicircular s emicircular then the time required for the electron moves from A to B is given by
t = t =
1 2 1 2
T
( 2.51× 10− )
t = 1.26 × 10 −7 s
7
PHYSICS
CHAPTER 6
Exercise 6.3 : 1.
Dete Determ rmin ine e the the sign sign of a cha charg rge e in in the the foll follow owin ing g probl pro ems. s. blem a.
F
B
b.
B
v
F
v
ANS. : positive; positive 2. Dete Determ rmin ine e the the direc directi tion on of the the mag magne neti tic c for force ce exer exerte ted d on on a positive charge in each problem below when a switch S is closed. a.
b.
+ Switch, S
v
v
Switch, S
+
PHYSICS
CHAPTER 6
Exercise 6.3 : 3.
An elect electro ron n expe experi rien ence ces s the the grea greate test st forc force e as it trav travel els s 2.9× 106 m s−1 in a magnetic field when it is moving north. The force is upward and of magnitude 7.2 × 10−13 N. Determine the magnitude and direction of the magnetic field. (Given the charge of the electron is 1.60 × 10−19 C) (Physics for scientists & engineers ,3 rd edition, Giancoli, Q22, p.705)
ANS. : 1.6 T to the east 4. An elec electtron ron movin oving g wit with a spee speed d of 9.1 9.1 × 105 m s−1 in the positive x direction experiences zero magnetic force. When it moves in the positive y direction, it experiences a force of 2.0× 10−13 N that points in the negative z direction. What is the direction and magnitude of the magnetic field? (Given e=1.60× 10−19 C and me= 9.11× 10−31 kg) (Physics, 3rd edition, James S. Walker, Q8, p.762)
ANS. : 1.37 T to the left (in the negative y direction)
PHYSICS
CHAPTER 6
Exercise 6.3 : 5.
Two Two char charge ged d part partic icle les s with with dif diffe fere rent nt spe speed eds s move move one one at at a time through a region of uniform magnetic field. The particles move in the same direction and experience equal magnetic forces. a. If particle particle 1 has four times times the the charge charge of particle particle 2, which which particle has the greater speed? Explain. b. Calcul Calculate ate the ratio ratio of of the the speed speeds, s, v1/v2. (Physics, 3rd edition, James S. Walker, Q9, p.762)
ANS. : 1/4 6. A 12.5 µ C particle of mass 2.80 × 10−5 kg moves perpendicular to a 1.01 T magnetic field in a circular path of radius 26.8 m. a. How fast fast is is the the parti particle cle moving moving? ? b. How long long will will it take take the the particle particle to to complete complete one one orbit? orbit? (Physics, 3rd edition, James S. Walker, Q18, p.763)
ANS. : 12.1 m s 1; 13.9 s
PHYSICS CHAPTER 6 Learning Outcome: s c i s y h p / y m . .. u d e . . ki r t a m . . h p m . .k w w w
6.4 Force Force on a curre currentnt-ca carry rrying ing cond conduc uctor tor in in a uniform magnetic field ( 1 hour) At the end of this cchapter, hapter, students should be able to: this chapter, Use formulae: F = I l × B
(
)
PHYSICS
CHAPTER 6
6.4 Force on a current-carrying conductor in a uniform magnetic field
When a current-carrying conductor is conductor is placed in a magnetic field B, thus a magnetic force will acts on that conductor . conductor . The magnitude of the magnetic force exerts on the currentcarrying conductor is given by
F = IlB sin θ In vector form, form,
(
F = I l × B
)
(6.10)
(6.11)
where F : magnetic force
B : magnitude of the magnetic flux density I : current l : length of the conductor θ : angle betw betwee een n direction of I and and B
PHYSICS
CHAPTER 6
The direction of the magnetic force can be determined by using the Fleming’s left hand rule as shown in Figure 6.27.
F
B
Note: Thumb – direction of Force
I First finger – direction of Field – Second finger – – direction of Current Figure 6.27
From the equation (6.10), the magnetic force on the conductor has its maximum value when the conductor (and therefore the current) and the magnetic field are perpendicular (at perpendicular (at right angles) to each other then =90 (shown in Figure 6.28a).
PHYSICS
CHAPTER 6 F max = IlB sin 90 F max = IlB
θ = 90
I
B
Figure 6.28a the magnetic force on the conductor is zero when the conductor (and therefore the current) is parallel to the magnetic field then =0 (shown in Figure 6.28b).
B
θ = 0
I Note:
F = IlB sin 0
F = 0
Figure 6.28b
One tesla is defined as the magnetic flux density of a field in which a force of 1 newton aacts cts on a 1 metre length newton acts length of of a conductor which carrying a current of 1 ampere and is is perpendicular to the field.
PHYSICS
CHAPTER 6
Example 9 : Determine the direction of the magnetic force, exerted on a currentcarrying conductor in the following cases. a.
X
X
X
X
X
X
X
X
b.
I
B X
X
X
X
X
X
X
X
X
I
B X
X
X
X
X
X
X
Solution : For both cases, use Fleming’s left hand rule : b. a. X
(to the left) F X
B X
X
X
X
X
X
X
X
F X
X
I X
X
X
X X
B X
X
X
I X
X
X(to the right) X
PHYSICS
CHAPTER 6
Example 10 : A wire of 100 cm long is placed perpendicular perpendicular to the magnetic field of 1.20 Wb m −2. a. Calculate the magnitude of the force on the wire when a current of 15 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 150 cm. c. If the force on the wire in part (b) is 60 × 10−2 N and the current flows is 12 A, calculate the magnitude of magnetic field was supplied. Solution : l = 1.00 m; B a. Given I 15 A
=
= 1.20 Wb m −2 ; θ = 90
F = IlB sin θ
= (15) (1.00 ) (1.20 ) sin 90 F = 18 N
PHYSICS
CHAPTER 6
Solution :
=
=
b. Given I 15 A; l 1.50 m The magnitude of the magnetic force on the wire is given by
F = IlB sin θ
= (15) (1.50 ) (1.20 ) sin 90 F = 27 N −2 c. Given I = 12 A; l = 1.50 m; F = 60 × 10 N
The magnitude of the magnetic field is given by
F = IlB sin θ
60 × 10 −2
= (12) (1.50) B sin 90 B = 3.33 × 10 −2 T
PHYSICS
CHAPTER 6
Example 11 : A straight horizontal rod of mass 50 g and length 0.5 m is is placed in a uniform magnetic field of 0.2 T perpendicular p erpendicular to the rod. The force acting on the rod just balances the rod’s weight. a. Sketch a labelled diagram shows the directions of the current, magnetic field, weight and force. b. Calculate the current in the rod. (Given g = 9.81 m s −2) Solution : m = 50 × 10 a.
−3
g; l = 0.5 m; B = 0.2 T; θ = 90 F B I
m g b. Since the magnetic force acting on the rod just balances the rod’s
= mg = IlB sin θ
F IlB sin θ weight, therefore
(50 × 10− )9.81 = I ( 0.5) ( 0.2) sin 90
3
I
4 91 A
PHYSICS CHAPTER 6 Learning Outcome: s c i s y h p / y m . .. u d e . . ki r t a m . . h p m . .k w w w
6.5 Forces Forces betwe between en two paral parallel lel curre current-ca nt-carryin rrying g conductors (1 hour) At the end of this cchapter, hapter, students should be able to: this chapter, Derive force per unit length of two parallel currentcarrying conductors. Use formulae:
F l
=
μ0 I 1 I 2 2πd
Define one ampere.
PHYSICS CHAPTER 6 6.5 Forces between two parallel currentcarrying conductors 6.5 .5.1 .1 For orce ce per uni unitt le leng ngth th
Consider two identical straight conductors 1 and 2 carrying currents I 1 and I 2 with length l are placed parallel to each other as shown in Figure 6.29.
1
2
I 1
F 12
I 2
P
B2
Q
F 21
I 1
d
Figure 6.29
I 2
B1
PHYSICS
CHAPTER 6
The conductors are in vacuum and their separation is d .
The magnitude of the magnetic flux density, B1 at point P on the conductor 2 due to the current in the conductor 1 is given by
B1
=
µ 0 I 1
2π d
Direction : into the page
Conductor 2 carries a current I 2 and in the magnetic field B1 thus the conductor 2 will experiences a magnetic force, F 12.
F 12 is given by The magnitude of F F 12 I 2lB1 sin θ and
=
θ = 90
µ 0 I 1 = I 2l sin 90 2π d
F 12
=
µ 0 I 1 I 2l
2π d
PHYSICS
CHAPTER 6
F 21 is given by The magnitude of F F 21 I 1lB2 sin θ and
=
θ = 90
µ 0 I 2 = I 1l sin 90 2π d µ 0 I 1 I 2l F 21 =
Conclusion :
F 12
2π d
= F 21 = F =
µ 0 I 1 I 2l
(6.12)
2π d
and the type of the force is attractive. attractive . From the equation (6.12), thus the force per unit length is given by
F l
=
μ0 I 1 I 2 2πd
(6.13)
PHYSICS
If the direction of current in the conductor 2 is change to upside down as shown in Figure 6.30.
1
Note:
CHAPTER 6
The currents are in the same direction – 2 conductors attract each other. The currents are in opposite direction – 2 conductors repel each other.
2
I 1
I 2
P
Q
B1
F 12
B2
F 21 I 1
d
I 2
Figure 6.30 F 12 and F 21 can be determined by using the eq. The magnitude of F (6.12) and their direction can be determined by applying Fleming’s left hand rule. Conclusion : Type of the force is repulsive. repulsive.
PHYSICS
CHAPTER 6
Example 12 : Two long straight parallel wires are placed 0.25 m apart in a vacuum. Each wire carries a current curr ent of 2.4 A in the same direction. a. Sketch a labelled diagram to show clearly clear ly the direction of the force on each wire. b. Calculate the force per unit length between the wires. c. If the current in one of the wires is reduced to 0.64 A, calculate the current needed in the second wire to maintain the same force per unit length between the wires as in (b). (Given µ 0 = 4π × 10−7 T m A−1)
=
Solution : I 1 I 2 a. The diagram is
= 2.4 A; d = 0.25 m
I 1 F 21
1
F 12
d
I 2
2
PHYSICS
CHAPTER 6
Solution : I 1
= I 2 = 2.4 A; d = 0.25 m
b. The force per unit length between the wires is given by
F
=
l
c. Given I 1
μ0 I 1 I 2
F
2πd
l F l
= 0.64 A
( 4π ×10 7 )( 2.4 ) ( 2.4 ) = −
2π ( 0.25 )
= 4.6 ×10
−6
N m −1
Therefore the current needed in the second wire is
F
=
l 4.6 ×10 −6
=
μ0 I 1 I 2 2πd −7 (4π ×10 )( 0.64 ) I 2
2π ( 0.25 ) I 2 = 8.98 A
PHYSICS
CHAPTER 6
6.5.2 The am ampere
From the eq. (6.13), if two long straight parallel conductors are placed 1.0 m apart in a vacuum and carry equal currents of 1.0 A thus the force per unit length that each conductor exerts on each other is given by
F
=
l =
μ0 I 1 I 2 2πd −7 (4π ×10 )(1)(1)
2π (1) F −7 −1 = 2.0 ×10 N m l
The ampere is defined as the constant current, which flowing in each of two infinitely long parallel straight conductors of negligible of cross sectional area separated by a distance dis tance of 1.0 1.0 metre in in vacuum, would produce a distance force per unit length between the conductors of 2.0 10 7 N m 1
PHYSICS
CHAPTER 6
Exercise 6.4 : Given µ 0 = 4π × 10−7 T m A−1 1.
A ver verti tica call str strai aigh ghtt con condu duct ctor or Y of of len lengt gth h 0.5 0.5 m is is sit situa uate ted d in in a uniform horizontal magnetic field of 0.1 T. a. Sketch a labelled labelled diagram diagram to show the the directi directions ons of the current, field and force. b. Calculate Calculate the the force force on Y when a current current of 4 A is passed passed into it. c. Through Through what what angle angle must must Y be turned turned in in a vertic vertical al plane plane so that the force on Y is halved? (Advanced level physics, 7th edition, Nelkon&Parker, Q6, p.336)
ANS. : 0.2 N; 60 2. A cur curre rent nt-ca -carr rryi ying ng cond conduc ucto torr exp exper erie ience nces s no no mag magne neti tic c for force ce when it is placed in a uniform magnetic field. Explain the statement.
PHYSICS CHAPTER 6 Learning Outcome: s c i s y h p / y m . .. u d e . . ki r t a m . . h p m . .k w w w
6.6 6. 6 To Torq rque ue on a coi coill (1 hour) At the end of this cchapter, hapter, students should be able to: this chapter, Use formulae: τ = NI NI A × B
(
)
where N = number of turns Explain the working principles of a moving coil galvanometer. Explain the DC electrical measuring instruments.
PHYSICS
CHAPTER 6
6.6 Torque on a coil 6.6.1 Formula of of to torque
Consider a rectangular coil (loop) of wire with side lengths a and
b that it can turn about axis PQ. The coil is in a magnetic field of flux density B and the plane of the coil makes an angle θ with the direction of the magnetic field. A current I is flowing round the coil as shown in Figure 6.31.
PHYSICS
CHAPTER 6
F
B
B F I I 1 A B P φ θ
B
F 1
a
b
B I
Q
I
Figure 6.31a
F
PHYSICS
CHAPTER 6
B
F 1
θ
φ
B
F 1
B
A rotation
b
φ
b
Q
2
φ
2 b
2
sin φ rotation
sin φ
Figure 6.31b: side view From the Figure 6.31b, the magnitude of the force F 1 is given by
F 1 = IlB sin 90 and l = a F 1 = IaB
From the Figure 6.31a, the forces F lie along the axis PQ.
PHYSICS
CHAPTER 6
From the Figure 6.31a, the forces F lie along the axis PQ.
The resultant force on the coil is zero but the nett torque is not zero because the forces F 1 are perpendicular to the axis PQ as shown in Figure 6.31a.
The forces F 1 cause the coil to rotate in the clockwise direction about the axis PQ as shown in Figure 6.31b.
The magnitude of the nett torque about the axis PQ (refer to Figure 6.31b) is given by
b b τ = − F 1 sin φ − F 1 sin φ 2 2 b = −2 F 1 sin φ and F 1 = IaB 2 b = −2( IaB ) sin φ 2 = − IabB sin φ and ab = A(area of coil)
PHYSICS
CHAPTER 6 τ = IAB sin φ since
φ = 90 − θ thus τ = IAB sin (90 − θ ) τ = IAB cos θ
For a coil of N turns, the magnitude of the torque is given by
τ = NIAB sin φ
(6.14)
OR
τ = NIAB cos θ
(6.15)
where τ : torque on the coil
B : magnetic flux density I : current flows in the coil φ : angle between vector area A and B θ : angle between the plane of the coil and B N : number of turns (coils)
PHYSICS
CHAPTER 6
From the eq. (6.14), thus the formula of torque in the vector form is given by (6.16) τ = NI NI A × B
(
)
The torque is zero when when
= 90 or = 0 and is maximum = 90 as shown in Figures 6.32a and 6.32b.
= 0 or
A
B
θ = 90
φ=0
B
OR
θ = 0 Figure 6.32b
τ = 0
plane of the coil
τ = NIAB cos 90
A
Figure 6.32a
τ = NIAB sin 0
φ = 90
τ = NIAB sin 90 OR
τ = NIAB cos 0
τ
= NIAB
PHYSICS
CHAPTER 6
In a radial field, field, the plane of the coil is always parallel to the magnetic field for any orientation of the coil about the vertical axis as shown in Figure 6.33.
N
S
θ = 0 OR
φ = 90
radial field fixed soft coil iron cylinder Figure 6.33: Plan coil meter Plan view of moving coil Hence the torque on the coil in a radial field is always constant and maximum given by τ NIAB cos 0 τ NIAB sin 90 OR
=
τ = NIA NIAB B
=
maximum
Radial field is used in moving coil galvanometer.
PHYSICS
CHAPTER 6
Example 13 : A 50 turns rectangular coil with with sides 10 cm × 20 cm is placed vertically in a uniform horizontal magnetic field of magnitude 2.5 T. If the current flows in the coil is 7.3 A, determine the torque acting on the coil when the plane of the coil is a. perpendicular to the field, b. parallel to the field, c. at 75° to the field.
=
7.3 A = 2.5 T; I = 7.3
A = (10 ×10
)( 20 ×10− ) = 2.0 ×10−
Solution : N 50 turns; B The area of the coil is given by −2 a.
2
2
m2
B From the figure, θ = 90° and φ = 0° , thus θ = 90 the torque on the coil is A τ = NIA NIAB B cos cos θ OR τ = NIAB sin φ
= NIA NIAB B cos cos 90 τ = 0
NIAB sin sin 0 = NIAB
PHYSICS
CHAPTER 6
Solution : N b.
= 50 turns; B = 2.5 T; I = 7.3 7.3 A
A
B
φ = 90
From the figure, θ = 0° and φ = 90° , thus the torque on the coil is
τ = NIAB cos θ = ( 50)() ( 7.3) ( 2.0 ×10 −2 )( 2.5) cos 0 τ = 18.3 N m
B
c.
θ = 75 φ = 15
A
From the figure, θ = 75° and φ = 15°,thus the torque on the coil is
τ = NIAB cos θ −2 = ( 50)() ( 7.3) ( 2.0 ×10 )( 2.5) cos 75 τ = 4.72 N m
PHYSICS
CHAPTER 6
6.6. 6. 6.2 2 Mo Movi ving ng-c -coi oill gal galva vano nome mete ter r
A galvanometer consists of a coil of of wire suspended in the magnetic field of a permanent magnet. The coil is rectangular shape and consists of many turns of fine wire as shown in Figure 6.34.
Figure 6.34
PHYSICS
CHAPTER 6
When the current I flows through the coil, coil , the magnetic field exerts a torque on the coil as given by
τ = NIA NIAB B
This torque is opposed by a spring which exerts a torque, given by
τ s where
= kθ
τ
(6.17)
k : torsional constant θ : rotation angle of the coil in radian
The coil and pointer will rotate only to the point where the spring torque balances the torque due to magnetic field , τ τ s thus
= NIAB = kθ I =
kθ NAB
(6.18)
s
PHYSICS
CHAPTER 6
Example 14 : A rectangular coil of 10 cm × 4.0 cm in a galvanometer has 50 turns and a magnetic flux density of 5.0 × 10−2 T. The resistance of the coil is 40 Ω and a potential difference of 12 V is applied across the galvanometer, calculate the maximum torque on the coil. −2 Solution : N 50 turns; B 5.0 10 T; R 40 ;
= V = 12 V
=
The area of the coil is given by −2
A = (10 ×10
×
=
)( 4.0 ×10− ) = 4.0 ×10− 2
3
Ω
m2
The current through the galvanometer is
V = IR
12 = I ( 40) I = 0.3 A
Therefore the maximum torque on the coil is
NIA NIAB B τ max max = = ( 50)() ( 0.3) ( 4.0 ×10−3 ) ( 5.0 ×10−2 ) 3 0 × 10 −3 N m τ
PHYSICS
CHAPTER 6
6.6. 6. 6.3 3 El Elec ectr tric ical al in inst stru rume ment nts s Ohmmeter It is used to measure the unknown resistance of the resistor . resistor . Figure 6.35 shows the internal connection c onnection of an Ohmmeter.
∞
0
where
RM : meter (coil) resistance RS : variable resistance
RX : unknown resistance
Ω
ε
RM
P
RX Figure 6.35
RS
Q
PHYSICS
CHAPTER 6
When nothing is connected to terminals P and Q , so that the circuit is open (that is, when R ), there is no current and no deflection. deflection. When terminals P and Q are short circuited (that is when 0), the ohmmeter deflects full-scale. full-scale . R = 0),
RX the meter deflection depends on the value of For any value of R RX.
Ammeter It is used to measure a current flows in the circuit . Ammeter is connected in series with other elements in the circuit because the current to be measured must pass directly through the ammeter .
RM) so that An ammeter should have low internal resistance ( R the current in the circuit would not affected. affected .
PHYSICS
CHAPTER 6
The maximum reading from the ammeter is known as full scale deflection (fs). If the full scale current passing through the ammeter then the potential difference (p.d.) across that ammeter is given by
V fs
= I fs RM
where RM : meter(coil) resistance
I fs : full scale current V fs : full scale potential potent ial difference (p.d.)
If the meter is used to measure currents that are larger than its full scale deflection ( I >I fs ), some modification has to be done.
A resistor has to be connected in parallel with the meter (coil) resistance RM so that some of the current will bypasses the meter (coil) resistance resistance..
This parallel resistor is called a shunt denoted as RS.
Figure 6.36 shows the internal connection of an ammeter with a shunt in parallel.
PHYSICS
CHAPTER 6 0
I
max
I fs
I S
A
I
RM RS
Figure 6.36 Since shunt is connected in parallel with the meter (coil) resistance then
= V R I fs RM = I S RS and I S = I − I fs I fs RM = ( I − I fs ) RS V RM
S
I fs RS = RM I − I fs
(6.19)
PHYSICS
CHAPTER 6
Voltmeter It is used to measure a potential difference (p.d.) across electrical elements in the circuit. circuit. Voltmeter is connected in parallel with other elements in the circuit therefore its resistance must be larger than the resistance of the element so that a very small amount of current only can flows through it. it. An ideal voltmeter has voltmeter has infinity resistance so that no current exist in it. To measure a potential difference that are larger than its full scale deflection (V > V fs ), the voltmeter has to be modified.
A resistor has to be connected in series with the meter (coil) resistance RM so that only a fraction of the total p.d. appears across the RM and the remainder appears across the serial resistor. This serial resistor is resistor is called a multiplier multiplier OR OR bobbin denoted as RB.
PHYSICS
CHAPTER 6
Figure 6.37 shows the internal connection of a voltmeter with multiplier in series.
0 V
I fs RB
I
I 1
max
RM V
Electrical element Figure 6.37 Since the multiplier is connected in series with the meter (coil) resistance then the current through them are the same, I fs .
PHYSICS
CHAPTER 6
The p.d. across the electrical element is given by
V = V RB
+ V R
M
Hence the multiplier resistance is
V = I fs RB + I fs RM
Note:
V − I fs RM RB = I fs
(6.20)
To convert a galvanometer to ammeter , a shunt (parallel resistor) is used. To convert a galvanometer to voltmeter , a multiplier (serial resistor) is used.
PHYSICS
CHAPTER 6
Example 15 : A milliammeter with a full scale deflection of 20 mA and an internal resistance of 40 Ω is to be used as an ammeter with a full scale deflection of 500 mA. Calculate the resistance of the shunt −3 −3 required. I 20 10 A; R 40 ; I 500 500 10 A fs
=
×
M
=
Ω =
Solution : By applying the formula of shunt resistor, thus
×
I fs RS = RM I − I fs 20 × 10 −3 = −3 −3 ( 40 ) 500 × 10 − 20 × 10 RS = 1.67 Ω
PHYSICS
CHAPTER 6
Example 16 : A galvanometer has an internal resistance of 30 Ω and deflects full scale for a 50 µ A current. Describe how to use this this galvanometer to make a. an ammeter to read currents up to 30 A. b. a voltmeter to give a full scale deflection of 1000 V. −6 Solution : I fs 50 10 A; RM 30
=
×
=
Ω
a. We make an ammeter by putting a resistor in parallel ( RS) with the internal resistance, RM of the galvanometer as shown in figure below.
I I fs
RM
I S
G RS
PHYSICS
CHAPTER 6
−6 Solution : I fs 50 10 A; RM a. Given I 30 A
=
=
×
= 30 Ω
Since RS in parallel with RM therefore
= V R = I S RS and I S = I − I fs I fs RM = ( I − I fs ) RS (50 × 10 −6 )( 30) = (30 − 50 × 10 −6 ) RS RS = 5.0 × 10 −5 Ω in parallel. V RM I fs RM
S
b. We make a voltmeter by putting a resistor in series ( RB) with the internal resistance, RM of the galvanometer as shown in figure R
RB
below.
I fs
M
G V
I fs
PHYSICS
CHAPTER 6
−6 Solution : I fs 50 10 A; RM b. Given V 1000 V
=
=
×
= 30 Ω
Since RB in series with RM therefore
V = V RB
+ V R V = I fs RB + I fs RM 1000 = (50 × 10 −6 ) RB + ( 50 × 10 −6 )( 30) RB = 2.0 × 10 7 Ω in series. M
PHYSICS
CHAPTER 6
Exercise 6.5 : 1.
A mov movin ing g coi coill met meter er has has a 50 turn turns s coi coill mea measu suri ring ng 1.0 1.0 cm cm by by 2.0 cm. It is held in a radial magnetic field of flux density 0.15 T and its suspension has a torsional constant of 3.0 × 10−6 N m rad−1. Determine the current is required to give a deflection of 0.5 rad. ANS. : 1.0 10 3 A
2.
A mil milliliam amme mete terr of of neg negliligi gibl ble e res resis ista tanc nce e pro produ duce ces s a full full sc scal ale e deflection when the current is 1 mA. How would you convert the milliammeter to a voltmeter with full scale deflection of 10 V? ANS. : 1.0 104 in series 3. A mov movin ingg-co coil il met meter has has a res resis ista tanc nce e of of 5. 5.0 Ω and full scale deflection is produced by a current of 1.0 mA. How can this meter be adapted for use as : a. a voltmeter reading up to 10 V, b. a ammeter reading up to 2? ANS. : 9995
in series; 2.5 10 3
in parallel
PHYSICS CHAPTER 6 Learning Outcome: s c i s y h p / y m . .. u d e . . ki r t a m . . h p m . .k w w w
6.7 Motion Motion of charg charged ed partic particle le in in magneti magnetic c field field and electric field ( 1 hour) At the end of this cchapter, hapter, students should be able to: this chapter, Explain the motion of a charged particle in both magnetic field and electric field. Derive and use formulae
v=
E B
in a velocity selector.
PHYSICS CHAPTER 6 6.7 Motion of charged particle in magnetic field and electric field
Consider a positively charged particle with mass m, charge q and velocity v enters a region of space where the electric and magnetic fields are perpendicular to the particle’s velocity v elocity and to each other as shown in Figure 6.38.
+ + + + + + + + F ++ + + + + + ++ +
+
v
X
X
X
X
X
X
X X
X
X E X
X
B
X
X
X
X
X
X
X
X
X
X
+
X X
v
− − − − − − − − F − E − − − − − − −− − Figure 6.38
B
+
v
PHYSICS
CHAPTER 6
The charged particle will experiences the electric force F E is downwards with magnitude qE and the magnetic force F B is upwards with magnitude qvB as shown in Figure 6.38.
If the particle travels in a straight line with a constant velocity hence the electric and magnetic forces are equal in magnitude.. Therefore magnitude
F B
qvB sin 90
= F E = qE
v=
E B
(6.21)
Only the particles with velocity equal to E/B can pass through without being deflected by the fields. fields. Eq. (6.21) also works for electron or other negatively charged particles.
PHYSICS
CHAPTER 6
Figure 6.38 known as velocity selector . Normally, after the charged particle passing through the velocity selector it will enter the next region consist of a uniform magnetic field only. only . This apparatus known as mass spectrometer as spectrometer as shown in Figure 6.39. X
X
X
X
r
X
X
X
X
+X
X
X
X
X
X B X B X F
X
X
X
X
X
X
X
vX
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
X
X
X
B
X
v
X
X
X
X
X
X
X
E
+
v
− − − − − − − − F − E − − − − − − −− − Figure 6.39
+
v
PHYSICS
CHAPTER 6
When the charged particle entering the region consist of magnetic field only, only , the particle will make a semicircular path of radius r as shown in Figure 6.39.Therefore
= F C 2 mv qvB = F B
q m
q m
=
=
r v
rB
E rB
2
and
v=
E B (6.22)
From the eq. (6.22), the mass spectrometer can c an be used to determine the value of q/ m for any charged particle.
PHYSICS
CHAPTER 6
Example 17 : An electron with kinetic energy of of 8.0 × 10−16 J passes perpendicular through a uniform magnetic field of 0.40 × 10−3 T. It is found to follow a circular path. Calculate a. the radius of the circular path. b. the time required for the electron e lectron to complete one revolution. (Given e/m = 1.76× 1011 C kg-1, me = 9.11× 10−31 kg) −16 −3 J; B 0.40 10 T Solution : K 8.0 10 a. The speed of the electron is given by
=
×
K =
8.0 × 10 −16
=
1
2 1
×
mv 2
= (9.11 × 10 −31 )v 2 2
v = 4.19 × 10 7 m s −1
PHYSICS
CHAPTER 6
−16 −3 J; B 0.40 10 T Solution : K 8.0 10 a. Since the path made by the electron is circular, thus
=
×
=
F B
evB sin 90
×
= F C 2 mv = r
e B = v m r 7 4 . 19 × 10 (1.76 ×1011 )( 0.40 × 10−3 ) = r
r = 0.595 m b. The time required for the electron e lectron to complete one revolution is given by
v= 4.19 × 10 7
=
2πr T 2π ( 0.595) T
T = 8.92 × 10 −8 s
PHYSICS
CHAPTER 6
Exercise 6.6 : 1.
An elec electtron ron mov movin ing g at at a st stead eady spe speed ed of 0.5 0.50 0 × 106 m s−1 passes between two flat, parallel metal plates 2.0 cm apart with a potential difference of 100 V between be tween them. The electron is kept travelling in a straight line perpendicular to the electric field between the plates by applying a magnetic field perpendicular to the electron’s path and to the electric field. Calculate : a. the intensity of the electric field.
b. the magnetic flux density needed. ANS. : 0.50 104 V m 1; 0.010 T 2.
A pro proto ton n mov movin ing g in in a circ circul ular ar path path perp perpen endi dicu cula larr to to a cons consta tant nt magnetic field takes 1.00 µ s to complete one revolution. Determine the magnitude of the magnetic field. (Physics for scientist and engineers, 6 th edition, Serway&Jewet, Q32, p.921)
(m p=1.67× 10−27 kg and charge of the proton, q=1.60× 10−19 C)
PHYSICS
CHAPTER 6
Next Chapter… CHAPTER 7 : Electromagnetic induction
