Download Force between two parallel wires is..

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Ch.Rama Krishna
Dr.Rk's frames of physics
Force between two parallel wires is..
when r >> R, B =
ELECTROMAGNETISM
In this chapter the student must concentrate on four important topics. i.e.
i) Calculation of magnetic field B
due to different configuration of
current elements which are the sources of magnetic field using Biot
Savart Law and Ampere's law.
ii) Finding the magnetic force on a
charged particle and predicting
the nature of path and Lorentz
force.
iii) Magnetic force on a current
carrying wire of different shapes.
iv) Torque and potential energy of
current loop in external magnetic
field.
1. Biot-Savart Law: Biot-Savart's
law is the magnetic analogue of
coulomb's law in Electrostatics.
The charge element dq appearing
in Coulomb's law is a scalar but
the current element idl appearing
inBiot-Savart's law is a vector
µ 0 2M
4π r 3
µ i d × rˆ
dB = 0
4π r 2
∫
µ ni
B = 0 {sin α + sin β}
2
where n is number of turns per
unit length. If solenoid is very
long α = β = 90°, Β = µ0ni
4. Force on a moving charge
in a
uniform magnetic field B :
The force on a charge q moving
with a velocity v in a magnetic
field B is given by F = qv × B .
i.e. F = qvB sin θ, where θ is
angle between v and B .
Nature of path followed by a
charged particle in magnetic field:
❖ If θ = 0° or 180°, F = 0, the path
is straight line.
❖ If θ = 90°, F = Bqv, the path is
circle of radius.
r=
The total field B = dB
where r̂ is unit vector in the dir
ection of vector r .
The direction of magnetic field
can be obtained from the right hand
thumb rule.
Magnetic field of a moving
charge
❖
(cyclotron frequency).
If θ lies between 0° and 90°, the
charged particle describes a helical path of radius mv sinθ/qB;
and pitch
θ=
1. θ = 0°
θ = 0°
Straight line path Straight line path
2.
θ = 90°
θ = 90°
When particle is projected outside the field and length of the
magnetic field is enough, the
angle with which the charged particle emerges out will be equal
to the angle with which it enters
into the field.
µ 0i
2 πr
Field due to a circular current carrying coil of radius R is
When a charged body having
charge q and mass m rotates with
an angular velocity ω about its
axis, the ratio of its magnetic
moment and angular momentum
M
q
=
L 2m
Torque on a current loop:
The torque experienced by a
current carrying
loop in a mag
B.
neticfield
Circular path
3. 0° < θ < 90°
0° < θ < 90°
B ⋅ d = µ 0 Ienclosed
closed loop
∫ B ⋅ d = µ 0 (I1 + I3 − I 2 )
c)
 µ I 
k  0  find the value of k.
 48πx 
µ0I
[cos 53° + cos 37 °]
12x
4π
5
 µ I 
= 7 0 
 48πx 
∴k=7
2. A magnetic field B = B0ˆj exists
in
the region a < x < 2a and
B = − B0ˆj in the region 2a< x <
3a, where B0 is a positive constant. A positive point charge
moving with a velocity v = v0ˆi
where v0 is a positive constant,
enters the magnetic field at x =
a. The trajectory of the charge
in this region can be like,
a)
b)
µ0iR 2 N
∫
Examples
1. A steady current 'I' goes through
a wire loop PQR having shape
of a right angle triangle with PQ
= 3x, PR = 4x and QR = 5x. If
the magnitude of the magnetic
field at P due to this loop is
dF = i dl × B; F = i l × B
Magnetic force on any closed
loop in a uniform field acting
perpendicular to plane of loop is
always zero.
F = 0.
∫ B ⋅ d) is equal to µ0 times the
U = −M ⋅ B
Force on magnetic dipole is nonuniform magnetic field.
Parabolic path Helical path
F = i PQ × B
(
Where M is magnetic dipole moment = niA.
Potential energy of a magnetic
dipole in a magnetic field is
6. Magnetic force on current
carrying conductor
Force
on
conductor:
curved
Ampere's Law:
Similar to the Gauss's law of
electrostatics, this law provides
us shortcut methods of finding
magnetic field in cases of symmetry. According to this law, the
line integral of magnetic field
over the closed path
τ = M ×B
B=
Parabolic path
d F = − (M ⋅ B)
dx
net current crossing the area
enclosed by the path.
Analysis:
Bq
t
m
x
θ = sin −1  
r
❖
2(R 2 + r 2 ) 3/ 2
5. Lorentz Force:
❖ The force experienced by a
moving charge with velocity v in
E
electric
a magnetic
field and
field B, F = q{E + v × B} .
❖ When E is parallel to B and particle velocity is perpendicular to
both of these fields, the path of a
particle is a helix with increasing
pitch.
❖ When electric field E is perpen
dicular to B and the particle is
released at rest from origin, the
path of a particle is cycloid.
Comparison
of paths in E field and B -field:
v cos θ2πm
Bq
in terms of length of the magnetic field x
µ i
B = 0 (sin α + sin β )
4ðr
B=
θ
T and PQ = 2r sinθ
π
After time t, the deviation will be
2. Magnetic field due to current
carrying conductor of finite
length:
α = β = 90°, B =
t=
qB
mv
and frequency f =
Bq
2 πm
x=
µ qv × rˆ
B= 0
4π r 2
If length of the conductor is
infinite or very long,
Sr. Inter
Physics
Physics
A current carrying loop acts like
magnetic
dipole of moment
M = NiA, where A is area of the
coil.
3. Magnetic field along the axis of
solenoid
Force between two parallel wires
carrying current is given by
F µ 0i1i 2
=
2πr
where M = Ni
πR2 = Magnetic moment
of the loop,
N = Number of turns.
i d
This law isused to find the
magnetic field dB at a point whose
position vector with
respect to a
current element i d is r . According
to this law: ❖
d)
Analysis:
for a < x < 2a
path will be concave upward
2a < x < 3a
path will be concave downward; So (a) is correct.
Ans: a
3. A thin flexible wire of length L
is connected to two adjacent
fixed points and carries a current I in the clockwise direction,
as shown in the figure. When
the system is put in a uniform
magnetic field of strength B going into the plane of the paper,
the wire takes the shape of a
circle. The tension in the wire is
[IIT-2010]
a) IBL
b)
IBL
c) 2π
d)
IBL
π
IBL
4π
Ans: c
The time spent in magnetic field