Chapter 11 Moving Charge and Magnetism ( Magnetic effects of Current Part 1) – Physics free study material by TEACHING CARE online tuition and coaching classes
Chapter 11 Moving Charge and Magnetism ( Magnetic effects of Current Part 1) – Physics free study material by TEACHING CARE online tuition and coaching classes
Biot-Savart’s law is used to determine the magnetic field at any point due to a current carrying conductors.
This law is although for infinitesimally small conductors yet it can be used for long conductors. In order to understand the Biot-Savart’s law, we need to understand the term current-element.
In the figure shown below, there is a segment of current carrying wire and P is a point where magnetic field is
to be calculated.
i dl
is a current element and r is the distance of the point ‘P’ with respect to the current element
i dl . According to Biot-Savart Law, magnetic field at point ‘P’ due to the current element
i dl
is given by the
expression,
dB = k i dlsinθ
also B =
dB = m 0 i .
dl sinq
r 2
In C.G.S. : k = 1 Þ dB =
ò
idl sinq
r 2
4p ò r 2
Gauss
In S.I. :
k = m0
Þ dB = m0 × idl sinq
Tesla
4p 4p r 2
where
m 0 = Absolute permeability of air or vacuum
= 4p ´ 10 -7 Wb . It’s other units are
Amp – metre
Henry metre
or N or
Amp 2
Tesla – metre Ampere
(1)
Different forms of Biot-Savarts law
- Similarities and differences between Biot-Savart law and Coulomb’s Law
(i) The current element produces a magnetic field, whereas a point charge produces an electric field.
- The magnitude of magnetic field varies as the inverse square of the distance from the current element, as does the electric field due to a point
dB = m 0
idl ´ rˆ
Biot-Savart Law
F = 1 q1q2 rˆ
Coulomb’s Law
4p r 2 4pe 0 r 2
(iii) The electric field created by a point charge is radial, but the magnetic field created by a current element is
perpendicular to both the length element dl and the unit vector rˆ.
E
+q
Direction of Magnetic Field.
The direction of magnetic field is determined with the help of the following simple laws :
- Maxwell’s cork screw rule
According to this rule, if we imagine a right handed screw placed along the current carrying linear conductor, be rotated such that the screw moves in the direction of flow of current, then the direction of rotation of the thumb gives the direction of magnetic lines of force.
(2) Right hand thumb rule
According to this rule if a current carrying conductor is held in the right hand such that the thumb of the hand represents the direction of current flow, then the direction of folding fingers will represent the direction of magnetic lines of force.
(3) Right hand thumb rule of circular currents
According to this rule if the direction of current in circular conducting coil is in
the direction of folding fingers of right hand, then the direction of magnetic field will be in the direction of stretched thumb.
(4) Right hand palm rule
If we stretch our right hand such that fingers point towards the point. At which magnetic field is required while thumb is in the direction of current then normal to the palm will show the direction of magnetic field.
Note : @ If magnetic field is directed perpendicular and into the plane of the paper it is represented by Ä (cross) while if magnetic field is directed perpendicular and out of the plane of the paper it is represented by ¤ (dot)
i i
CW ACW
In Out
In : Magnetic field is away from the observer or perpendicular inwards.
Out : Magnetic field is towards the observer or perpendicular outwards.
Application of Biot-Savarts Law.
(1) Magnetic field due to a circular current
If a coil of radius r, carrying current i then magnetic field on it’s axis at a distance x from its centre given by
Baxis
= μ0 .
4π
2πNir 2
(x 2 + r 2 )3/2
; where N = number of turns in coil.
Different cases
Case 1 : Magnetic field at the centre of the coil
(i) At centre x = 0 Þ B
= m0 . 2pNi = m 0 Ni = B
centre 4p r 2r max
(ii) For single turn coil N = 1 Þ B
= m0 . 2pi = m0i
(iii) In C.G.S.
m0 = 1 Þ B
= 2pi
centre
4p r 2r
4p centre r
Note : @ B
centre µ N
(i, r constant),
Bcentre µ i
(N, r constant),
Bcentre µ r
(N, i constant)
|
Case 2 : Ratio of Bcentre and Baxis
B æ x 2 ö3/2
|
|
The ratio of magnetic field at the centre of circular coil and on it’s axis is given by centre = ç1 + ÷
a 5 5
Baxis ç
a æ 3 ö3 / 2
r 2 ÷
(i) If
x = ±a, Bc = 2 Ba
x = ± 2 , Bc = 8 Ba
x = ± , Bc = ç ÷ Ba
2 2
- If
B = Bc
a n
then
x == ±r
and if Ba = c
è ø
|
then x == ±r
Case 3 : Magnetic field at very large/very small distance from the centre
(i) If x >> r (very large distance) Þ B
= m0 . 2p Nir 2
= m0 . 2NiA
where A = pr2 = Area of each turn of the coil.
axis 4p x 3
4p x 3
(ii) If x << r (very small distance) Þ
Baxis
¹ Bcentre , but by using binomial theorem and neglecting higher
x 2 æ 3 x 2 ö
power of
r 2 ; Baxis = Bcentre ç1 – 2 r 2 ÷
è ø
Case 4 : B–x curve
The variation of magnetic field due to a circular coil as the distance x varies as shown in the figure.
B varies non-linearly with distance x as shown in figure and is maximum when at the centre of the coil and it is zero at x = ± ¥.
x 2 = min = 0 , i.e., the point is
Point of inflection (A and A¢) : Also known as points of curvature change or pints of zero curvature.
(i) At these points B varies linearly with x Þ
dB =
dx
constant Þ
d 2 B = 0 .
dx 2
(ii) They locates at
x = ± r
2
from the centre of the coil.
(iii) Separation between point of inflextion is equal to radius of coil (r)
- Application of points of inflextion is “Hamholtz coils”
Note : @The magnetic field at
(2) Helmholtz coils
x = r
2
is B = 4m0 Ni
(i) This is the set-up of two coaxial coils of same radius such that distance between their centres is equal to their radius.
- These coils are used to obtain uniform magnetic field of short range which is obtained between the
(iii) At axial mid point O, magnetic field is given by
B = 8m0 Ni = 0.716 m0 Ni = 1.432 B , where
B = m0 Ni
5 5R R 2R
(iv) Current direction is same in both coils otherwise this arrangement is not called Helmholtz’s coil arrangement.
- Number of points of inflextion Þ Three (A, A¢, A¢¢)
Note : @The device whose working principle based on this arrangement and in which uniform magnetic field is used called as “Halmholtz galvanometer”.
- Magnetic field due to current carrying circular arc : Magnetic field at centre O
B = m0 . pi = m0i
B = m0 .q i
B = m0 . (2p – q )i
4p r 4r
4p r
4p r
Special results
If magnetic field at the centre of circular coil
is denoted by B
æ= m0 . 2pi ö
0 ç 4p r ÷
|
|
Magnetic field at the centre of arc which is
making an angle q at the centre is
Barc
= æ B0 ö.q
|
2p
è ø
- Concentric circular loops (N = 1)
(i) Coplanar and concentric : It means both coils are in same plane with common centre
(a) Current in same direction (b) Current in opposite direction
i
|
i
|
B = 0
4p
æ
|
2pi ç
è r1
ö
|
+ ÷
r2 ø
B = 0
|
|
4p
2pié 1
|
ë
1 ù
|
|
r2 ú
B1 æ r2 + r1 ö
Note : @
B = ç r – r ÷
2 è 2 1 ø
(ii) Non-coplanar and concentric : Plane of both coils are perpendicular to each other
Magnetic field at common centre
m
B = = 0
2r
(5) Magnetic field due to a straight current carrying wire
Magnetic field due to a current carrying wire at a point P which lies at a perpendicular distance r from the wire as shown is given as
|
m i
. (sin
- sinf )
4p r 1 2
From figure a = (90o – f1) and
m
b = (90o + f2 )
Hence B = o . i (cosa – cos b )
4p r
Different cases
Note : @ When point P lies on axial position of current carrying conductor then magnetic field at P
B = 0 i
P
@ The value of magnetic field induction at a point, on the centre of separation of two linear parallel conductors carrying equal currents in the same direction is zero.
- Zero magnetic field : If in a symmetrical geometry, current enters from one end and exists from the other, then magnetic field at the centre is
i i
In all cases at centre B = 0
Example: 1 Current flows due north in a horizontal transmission line. Magnetic field at a point P vertically above it directed
- North wards
- South wards
- Toward east
- Towards west
Solution : (c) By using right hand thumb rule or any other rule which helps to determine the direction of magnetic field.
Example: 2 Magnetic field due to a current carrying loop or a coil at a distant axial point P is B1 and at an equal distance
in it’s plane is B2
then B1 is
B
2
(a) 2 (b) 1 (c)
1 (d) None of these
2
Solution : (a) Current carrying coil behaves as a bar magnet as shown in figure.
We also knows for a bar magnet, if axial and equatorial distance are same then Ba = 2Be
Hence, in this equation
B1 = 2
B2 1 B1
Example: 3 Find the position of point from wire ‘B‘ where net magnetic field is zero due to following current distribution
- 4 cm
30 cm
7
12 cm
7
- 2 cm
Solution : (c) Suppose P is the point between the conductors where net magnetic field is zero.
So at P |Magnetic field due to conductor 1| = |Magnetic field due to conductor 2|
i.e.
m 0 . 2(5i) = m 0 . 2(2i) Þ
5 = 9
Þ x = 30 cm
4p i
4p (6 – x)
x 6 – x 7
Hence position from B = 6 – 30 = 12 cm
7 7
cm
Example: 4 Find out the magnitude of the magnetic field at point P due to following current distribution
(a)
m 0ia
pr 2
m ia 2
(b)
(c)
(d)
0
pr
m0ia
2pr 2
2m 0ia
pr 2
Solution : (a) Net magnetic field at P, B = 2B sinq ; where B = magnetic field due to one wire at P = m0 . 2i
net
and sinq = a \ B = 2 ´ m 0 . 2i ´ a = m 0 ia .
4p r
r net
4p r r
pr 2
Example: 5 What will be the resultant magnetic field at origin due to four infinite length wires. If each wire produces magnetic field ‘B‘ at origin
- 4 B
- B
- 2 B
- Zero
Solution : (c) Direction of magnetic field (B1, B2, B3 and B4) at origin due to wires 1, 2, 3 and 4 are shown in the following figure.
m 0 2i i 1 2
B1 = B2 = B3 = B4
= 4p . x
= B . So net magnetic field at origin O
i i
B4
Bnet =
=
= 2 2B
B2
O B3 B1
4 i
3
Example: 6 Two parallel, long wires carry currents i1 and i2 with i1 > i2 . When the currents are in the same direction, the
magnetic field at a point midway between the wires is 10 mT. If the direction of i2
becomes 30 mT. The ratio i1 / i2 is
is reversed, the field
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|
(a) 4 (b) 3 (c) 2 i i
Solution : (c) Initially when wires carry currents in the same direction as shown.
Magnetic field at mid point O due to wires 1 and 2 are respectively
B = m 0 . 2i1 Ä and B
= m0 . 2i2 ¤
1 4p x 2 4p x
Hence net magnetic field at O B
= m 0 ´ 2 (i – i )
Þ 10 ´ 10 –6 = m 0 2 – i )
net
4p x 1 2
…..(i)
4p . x (i1 2
i1 i2
If the direction of i2 is reversed then
B = m 0 . 2i1 Ä and B = m 0 . 2i2 Ä
1 4p x 2 4p x
So B
= m 0 2
- i )
Þ 30 ´ 10 –6 = m 0 2
+ i )…… (ii)
net
4p . x (i1 2
4p . x (i1 2
Dividing equation (ii) by (i)
i1 + i2 = 3 Þ i1 = 2
i1 – i2 1 i2 1
Example: 7 A wire of fixed length is turned to form a coil of one turn. It is again turned to form a coil of three turns. If in both cases same amount of current is passed, then the ratio of the intensities of magnetic field produced at the centre of a coil will be [MP PMT 2002]
- 9 times of first case (b)
1 times of first case (c) 3 times of first case (d)
9
1 times of first case
3
Solution : (a) Magnetic field at the centre of n turn coil carrying current i
B = m 0 . 2pni
……(i)
4p r
For single turn n =1
B = m 0 . 2pi
…..(ii)
4p r
If the same wire is turn again to form a coil of three turns i.e. n = 3 and radius of each turn r‘ = r
3
So new magnetic field at centre B‘ = m 0 . 2p (3)
Þ B‘ = 9 ´ m 0 . 2pi
……(iii)
4p r‘ 4p r
Comparing equation (ii) and (iii) gives B‘ = 9B .
Example: 8 A wire in the form of a square of side a carries a current i. Then the magnetic induction at the centre of the square wire is (Magnetic permeability of free space = m0) [EAMCET 2001]
(a)
m0 i
2pa
m0 i
pa
2 2m 0i
pa
m 0 i
i
Solution : (c) Magnetic field due to one side of the square at centre O
i
B = m 0 .
1 4p
2i sin 45 o Þ
a / 2
B = m 0 .
1 4p a
Hence magnetic field at centre due to all side B
net
= 4 B1
= m 0 (2 2 i)
pa
a/2
a
Example: 9 The ratio of the magnetic field at the centre of a current carrying circular wire and the magnetic field at the centre of a square coil made from the same length of wire will be
- p2
- p2
p (d) p
2 2 4 2
Solution : (b) Circular coil Square coil i i
i
i
Length L = 2p r Length L = 4a
Magnetic field B = m 0 . 2pi
= m 0 . 4p 2i
B = m 0 . 2
2 i B = m 0 . 8 2 i
Hence
4p r
Bcircular = p 2
Bsquare
4p r
4p a
4p a
Example: 10 Find magnetic field at centre O in each of the following figure
- (ii) (iii)
m 0i
m 0i æ 1 1 ö
m 0i æ 1 1 ö
(a)
r
Ä (a)
ç
4 è r1
– ÷ Ä (a)
r2 ø
ç
4 è r1
– ÷ Ä
r2 ø
m 0i
m 0i æ 1 1 ö
m 0i æ 1 1 ö
(b)
- (b)
2r
ç
4 è r1
+ ÷ Ä (b)
r2 ø
ç
4 è r1
+ ÷ Ä
r2 ø
m 0i
m 0i æ 1 1 ö
m 0i æ 1 1 ö
(c)
Ä (c)
4r
ç
4 è r1
– ÷ ¤ (c)
r2 ø
ç
4 è r1
– ÷ ¤
r2 ø
m 0 i ¤ (d) Zero (d) Zero
4r
Solution : (i) (c) Magnetic field at O due to parts 1 and 3, B1 = B3 = 0
(2)
While due to part (2)
B = m 0 . pi Ä
2 4p r
\ Net magnetic field at centre O,
(1)
O (3)
m p i m i 2
Bnet
= B1 + B2 + B3 = 0 .
|
4 r
Ä Þ Bnet
= 0 Ä
4r
- (b) B1 = B3 = 0
B = m 0 . p i Ä
2 4p r1
B = m 0 . p i Ä 4
4 4p r2
m 0 æ 1 1 ö
So Bnet
= B2 + B4
= .p iç
4p è r1
+ ÷ Ä
r2 ø
- (a) B1 = B3 = 0
B = m 0 . p i Ä
2 4p r1 4
B = m 0 . p i Ä
As | B
| > | B |
4 4p r2
2 4
m 0i æ 1 1 ö O
So Bnet
= B2 – B4 Þ Bnet =
ç
4 è r1
– ÷ Ä 1 3
r2 ø
Example: 11 Find magnetic field at centre O in each of each of the following figure
- (ii) (iii)
(a)
m 0 i ¤ (a)
2r
m 0 i (p – 2) Ä (a)
2p r
m 0 2i (p + 1) Ä 2r r
(b)
m 0 i Ä (b)
m 0i
i p + 2) ¤ (b)
m 0i
2i p – 1) Ä
(c)
2r
3m 0i Ä
8r
(c)
4p . r (
m 0i Ä
4r
4r . r (
- Zero
(d)
3m 0 i ¤ (d)
8r
m 0 i ¤ (d) Infinite
4r
Solution : (i) (d) By using B = m 0 . (2p – q )i
Þ B = m 0 . (2p – p / 2)i = 3m 0i ¤
4p r 4p r 8r
- (b) Magnetic field at centre O due to section 1, 2 and 3 are respectively
B = m 0 . i ¤
1 4p r 1 i
m pi
|
B = 0 . ¤
4p r
B = m 0 . i ¤
3 4p r
2
m 0 i
Þ Bnet
= B1 + B2 + B3 = 4p r (p + 2) ¤
- (b) The given figure is equivalent to following figure, magnetic field at O due to long wire (part 1)
B = m 0 . 2i ¤
1 4p r
m 2p i
|
Due to circular coil B = 0 . Ä 2
4p r
Hence net magnetic field at O
B = B
– B = m 0 2i p – 1) Ä
net
2 1 4p . r (
1
Example: 12 The field B at the centre of a circular coil of radius r is p times that due to a long straight wire at a distance r
from it, for equal currents here shows three cases; in all cases the circular part has radius r and straight ones are infinitely long. For same current the field B is the centre P in cases 1, 2, 3 has the ratio [CPMT 1989]
|
(1) (2) (3)
(a)
æ- p ö : æ p ö : æ 3p – 1 ö
(b)
– p + ö æ p +
ö æ 3p + 1 ö
ç 2 ÷ ç 2 ÷ ç 4 2 ÷
ç 2 1÷ : ç 2
1÷ : ç 4 2 ÷
è ø è ø è ø è ø è ø è ø
(c)
– p p 3p
(d)
æ- p – 1ö : æ p – 1 ö : æ 3p + 1 ö
2 : 2 : 4
ç ÷ ç
|
|
|
|
|
è ø è
÷ ç ÷
ø è ø
Solution : (a) Case 1 : B = m 0 . i Ä
(A)
A 4p r
B = m 0 . i ¤
B 4p r
B = m 0 . i ¤
C 4p r
(C)
So net magnetic field at the centre of case 1
B = B – (B + B )
Þ B = m 0 . pi ¤…………… (i)
1 B A C
1 4p r
Case 2 : As we discussed before magnetic field at the centre O in this case
(B)
B = m 0 . pi Ä
…..(ii)
2 4p r
Case 3 : BA = 0
- O
B = m 0 . (2p – p / 2) Ä = m 0 . 3pi Ä
B 4p r
B = m0 . i ¤
4p 2r
C 4p r
So net magnetic field at the centre of case 3
B = m 0 . i æ 3p – 1ö Ä
….(iii)
|
|
|
|
3 4p ç ÷
From equation (i), (ii) and (iii)
B : B : B
= p ¤ : p ¤ : æ 3p – 1ö Ä = – p
: p : æ 3p – 1 ö
|
|
1 2 3
ç ÷ ç ÷
|
|
è ø è 4 ø
Example: 13 Two infinite length wires carries currents 8A and 6A respectively and placed along X and Y-axis. Magnetic field at a point P (0, 0, d)m will be
(a)
7m 0
pd
(b)
10m 0
pd
(c)
14 m 0
pd
(d)
5m 0
pd
Solution : (d) Magnetic field at P
Due to wire 1,
B = m 0 . 2(8)
1 4p d
m 2(16)
and due to wire 2,
B = 0 .
|
4p d
m 2 5m
\ Bnet = =
= 0 ´
4p
´ 10 = 0
d pd
Example: 14 An equilateral triangle of side ‘a‘ carries a current i then find out the magnetic field at point P which is vertex of triangle
(a)
(b)
m 0i Ä
m 0i ¤
2 3pa
(c)
2 3m 0i ¤
pa
- Zero
Solution : (b) As shown in the following figure magnetic field at P due to side 1 and side 2 is zero.
Magnetic field at P is only due to side 3,
which is B1
= m 0 . 2i sin 30 o
|
|
4p 3a
2 2
= m 0 . 2i ¤ = m 0i ¤ 3
4p 3a a
Example: 15 A battery is connected between two points A and B on the circumference of a uniform conducting ring of radius r and resistance R. One of the arcs AB of the ring subtends an angle q at the centre. The value of, the magnetic induction at the centre due to the current in the ring is [IIT-JEE 1995]
(a) Proportional to 2(180 o – q )
- Zero, only if q = 180 o
(b) Inversely proportional to r
(d) Zero for all values of q
Solution : (d) Directions of currents in two parts are different, so directions of magnetic fields due to these currents are different.
Also applying Ohm’s law across AB
i1 R1 = i2 R2 Þ i1l1 = i2l 2 …..(i)
Also B
= m0 ´ i1l1
and B
= m0 ´ i2l2 ; \
B2 = i1l1 = 1 [Using (i)]
1 4p r 2
2 4p r 2
B1 i2l2
Hence, two field are equal but of opposite direction. So, resultant magnetic induction at the centre is zero and is independent of q .
Example: 16 The earth’s magnetic induction at a certain point is 7 ´ 10–5 Wb / m2 . This is to be annulled by the magnetic
induction at the centre of a circular conducting loop of radius 5 cm. The required current in the loop is
[MP PET 1999; AIIMS 2000]
(a) 0.56 A (b) 5.6 A (c) 0.28 A (d) 2.8 A
Solution : (b) According to the question, at centre of coil B = B
Þ m 0 . 2p i = B
Þ 10 -7 ´
2p i
(5 ´ 1-2 )
= 7 ´ 10 -5
Þ i = 5.6
H
amp.
4p r H
Example: 17 A particle carrying a charge equal to 100 times the charge on an electron is rotating per second in a circular
path of radius 0.8 metre. The value of the magnetic field produced at the centre will be (m0 –
permeability for
vacuum) [CPMT 1986]
(a)
10–7
m0
(b)
10–17 m0
(c)
10–6 m0
(d)
10–7 m0
Solution : (b) Magnetic field at the centre of orbit due to revolution of charge.
B = m 0 . 2p (qn ) ; where n = frequency of revolution of charge
4p r
So,
B = m 0 ´ 2p ´ (100e ´ 1)
Þ B = 10 –17 m .
4p 0.8 0
Example: 18 Ratio of magnetic field at the centre of a current carrying coil of radius R and at a distance of 3R on its axis is
(a)
10 (b)
20 (c)
2 (d)
|
B æ x 2 ö 3 / 2 B
Solution : (a) By using
centre = ç1 +
÷ ; where x = 3R and r = R Þ centre = (10)3 / 2 = 10
10 .
|
Baxis
ç r 2 ÷
Baxis
Example: 19 A circular current carrying coil has a radius R. The distance from the centre of the coil on the axis where the
magnetic induction will be
- th to its value at the centre of the coil, is [MP PMT 1997]
8
(a)
R (b)
R (c)
- 3 R
- R
|
B æ x 2 ö 3 / 2 1
Solution : (b) By using
centre = ç1 +
÷ , given r = R and Baxis =
Bcentre
|
Baxis
ç r 2 ÷ 8
|
|
|
|
|
æ 2 ö 3 / 2
ìæ 2 ö1 / 2 ü3
æ 2 ö1 / 2 2
Þ 8 = ç1 + x ÷
Þ (2)2 = ïç1 + x ÷ ï Þ 2 = ç1 + x ÷
Þ 4 = 1 + x
Þ x = 3R
|
|
ç R 2 ÷
íç ÷ ý
îè ïþ
ç R 2 ÷ R 2
Example: 20 An infinitely long conductor PQR is bent to form a right angle as shown. A current I flows through PQR. The
magnetic field due to this current at the point M is
H1. Now, another infinitely long straight conductor QS is
connected at Q so that the current is
1 in QR as well as in QS, the current in PQ remaining unchanged. The
2
magnetic field at M is now H2. The ratio H1 / H2 is given by [IIT-JEE (Screening) 1999]
- 12
– ¥ i
M
90o
Q + ¥
- 1
2
P 90o S
3 R
- 2
Solution : (c) Magnetic field at any point lying on the current carrying conductor is zero.
Here H1 = magnetic field at M due to current in PQ
H = magnetic field at M due to R + due to QS + due to PQ = 0 + H1 + H
2 2 1
= 3 H1
2
\ H1 = 2
H 2 3
Example: 21 Figure shows a square loop ABCD with edge length a. The resistance of the wire ABC is r and that of ADC is 2r. The value of magnetic field at the centre of the loop assuming uniform wire is
- 3pa ¤
- 3pa Ä
(c) p a ¤
(d) p a Ä
Solution : (b) According to question resistance of wire ADC is twice that of wire ABC. Hence current flows through ADC is
half that of ABC i.e.
i2 = 1 . Also i + i
= i Þ i = 2i and i = i
i1 2 1 2
1 3 2 3
Magnetic field at centre O due to wire AB and BC (part 1 and 2)
B = m 0 .
1 4p
2i1 sin 45 o Ä
a / 2
= m 0 . 4p
2 2 i1 Ä
a
and magnetic field at centre O due to wires AD and DC (i.e. part 3 and 4) B = B
= m 0 2 2 i2 ¤
Also i1 = 2i2. So (B1 = B2) > (B3 = B4)
Hence net magnetic field at centre O
3 4 4p a
B
Bnet
= (B1 + B2 ) – (B3 + B4 )
2 2 ´ æ 2 i ö 2 2 æ i ö ´ 2
m ç 3 ÷ m
ç 3 ÷ m
2 m i
= 2 ´ 0 . è ø – 0 . è ø = 0 .
(2 – 1) Ä = 0 Ä
4p a
4p a
4p 3a
3p a
Tricky example: 1
Figure shows a straight wire of length l current i. The magnitude of magnetic field produced by the
current at point P is l
P
i
l
(a)
2m 0i
pl
(b)
m 0 i
4pl
2m 0 i
8pl
m 0 i
2 2pl
Solution: (c) The given situation can be redrawn as follow.
As we know the general formula for finding the magnetic field due to a finite length wire
|
m i
. (sin
- sin f )
4p r 1 2 l
P
Here f1
= 0o, f = 45o
45o
i
\ B = m 0 . i (sin 0 o + sin 45 o ) = m 0 . i
Þ B = 2m 0i l
4p r
4p 2l
8pl
Tricky example: 2
A cell is connected between the points A and C of a circular conductor ABCD of centre ‘O‘ with angle
AOC = 60°, If B1 and B2 are the magnitudes of the magnetic fields at O due to the currents in ABC
and ADC respectively, the ratio B1 is
B2
(a) 0.2
- 6
i1 B
O
A
300o
60o
C
[KCET (Engg./ Med.) 1999]
Solution: (c)
- 1
- 5
B = m 0 . q i
i2 D
1A
4p r
Þ B µ q i
Þ B1 = q 1 ´ i1
B2 q 2 i2
Also i1
= l 2
= q 2
Hence
B1 = 1
i2 l1 q1 B2 1
admin
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