1) Ampere's circuital law states that the line integral of the magnetic field B around any closed path is equal to the permeability of free space times the total current passing through the enclosed area. 2) The law can be used to calculate magnetic fields due to various current carrying conductors like long straight wires, solenoids, and toroids. 3) For a long straight wire, the magnetic field at a distance r is given by B=μ0I/2πr. For a solenoid, the magnetic field inside is uniform and given by B=μ0nI, where n is number of turns per unit length. For a toroid, the magnetic field within is also

1. AMPERE’S CIRCUITAL LAW:--
The line integral of the magnetic field B around any closed path in free space
is equal to μo times the total current ( I ) passing through that closed path
∮B .dl = μo × I
Amperes law is applicable only for an Amperian loop as the Gausss law is used for Gaussian
surface in electrostatic.
The choice of an Amperian loop has to be such that, at each point of the loop either
1 ) B is tangential to the Amperian loop or B is a non zero constant
2) B is normal to the loop
3 ) B vanishes
Proof
Let us consider an infinitely long straight conductor carrying a current I .
From bio Savart law the magnitude of the magnetic field B due to current
carrying conductor at any point, distant are can be written as
The magnetic field lines are concentric circle perpendicular to the plane of the paper with
Central lyings on the conductor
μ0 2I
B =
2πa
ao

2. The field B at any point is directed along the tangent to the magnetic field lines at that point
shown in the figure.
The magnitude of the field B is same for all the points on the circle.
in order to evaluate the line integral of magnetic field B along the circle we considered as a
small current element dl along the circle.
At every point on the circle both B and dl tangential to the circle so that angle between them
is zero.
The line integral of the magnetic field along the circular path
∮B .dl = μo × I
∮B .dl = ∮B dl cos0°
= B ∮dl
= B.2 π a
× 2 πa
μ0 2I
4πa
=
Cos0 = 1
ao

3. dθ1
r1
dl1
Proof-2 (Irregular coil):
To prove:
∮B .dl = ∮B dl cos0°
Starting from the left hand side:
∮ B dl =
We know that: dθ1 = dl1/r1
∴ ∫B.dl1 = × [ dθ1 + dθ2 + dθ3 + dθ4 + dθ5 + …….]
∫B . dl =
= μo i
Cos0 = 1
θ = 0°
∮B .dl = μo × I
+
μoi
2πr2
× dl2 +
μoi
2π
∫ dθ = 2π
B.dl1 + B.dl2 +B.dl3 + B.dl4 + B.dl5 + B.dl6 + B.dl7 …….
μoi
2πr5
× dl5 …….+
μoi
2πr4
× dl4 +
μoi
2πr3
×
dl3
=
μoi
2πr1
× dl1
× 2π
μoi
2π

4. Note:
• Magnetic field at a point doesn’t depend on the shape of the Amperian loop.
• Magnetic field is same at every point in the Amperian loop (magnetic field possesses
cylindrical symmetry)
• Direction of magnetic field at any point on the Amperian loop is tangential to the circle formed
at that point with wire passing through the center.
• And the direction could be calculated by right hand thumb rule where, on holding the current
carrying wire such that the extended thumb shows the direction of current in the wire, then the
curling of rest of the 4 fingers represent the direction of rotation of magnetic field.
B
I

5. Applications of AMPERE’S CIRCUITAL LAW:--
Let us apply this law to compute magnetic field due to a long straight current carrying wire a
long straight, solenoid and endless (toroidal) solenoid.
ao
μ0 2I
B =
4πa
1. Magnetic field due to a long straight current carrying conductor :--
Consider a long straight wire x y carrying a current I.
P is a point at a perpendicular distance r from it at which magnetic field is required.
If B is magnetic field at P then it will be acting tangentially to the magnetic field line passing
through .
We consider an amperian circular loop of radius r with centre on wire such that point P lies on
the loop.
By symmetry the magnitude of field is same at every point of the circular loop and B and dl
are acting in the same direction .
∮B .dl = ∮B dl cos0°
= B ∮dl
= B.2πa ……….. 1
∮B .dl = μo × I ………….. 2

6. From equation (1) and (2)
B × 2πa = μo × I
μ0 I
B =
2πa
μ0 2I
B =
4πa
1 The above derivation is not valid for the magnetic field due to a current carrying straight
conductor of finite length . This is because a wire of finite length alone cannot form a
complete current circuit .It requires some additional wire to form a complete closed circuit.
However this difficulty does not arise in infinitely long wire.
2 From equation it is obvious that magnitude of the magnetic field at every point of a circle
of radius r with wires along the axis is same. It means that magnetic field due to current
through infinite strength wire has a cylindrical symmetry.
3 Though the wire carrying current is infinite, the magnetic field due to it at non zero
distance is not infinite. Since B ∝ 1/r so magnitude of magnetic field decrease as distance of
the point from the wire decreases.
B
I

7. 2 Magnetic field of a solenoid :---
A solenoid is a coil of many turns of an insulated copper wire closely wound in the shape of a tight spring.
A solenoid is a long wire wound in a close-packed helix carrying a current I and the length of the solenoid is
much greater then its diameter.
When an electric current flow through a solenoid ,magnetic field is set up around solenoid similar to that of a
bar magnet.
One end of a solenoid act as a north pole and other as south pole.
Magnetic field is represented by straight magnetic field lines parallel and very close to each other.
Magnetic field inside a long solenoid decreases as we move towards ends of solenoid because magnetic field
lines near the ends of solenoid start spreading out.
•The solenoid magnetic field is the vector sum of the field produced by the individual turns that make up the
solenoid
•Magnetic field B is nearly uniform and parallel to the axis of the solenoid at interior points near its center and
external field near the center is very small assumed to zero.
•Consider a dashed closed path abcd as shown in figure .Let l be the length of side ab of the loop which is
parallel to the is of the solenoid.
•At side a b magnetic field B is approximately parallel and constant. So for this side
∮B .dl = Bl

8. I I
xxxxx xx
TIP :
When we look at any end of the coil carrying current, if the current is in anti-clockwise direction
then that end of coil behaves like North Pole and if the current is in clockwise direction then that
end of the coil behaves like South Pole.
B
Let us also consider that sides bc and da of the loop are very-very long so that side cd is very
much far away from the solenoid and magnetic field at this side is negligibly small and for
simplicity we consider its equal to 0 .

9. ∮ B.dl = ∮Bdl cos0 + ∮Bdl cos90 + ∮Bdl cos0 + ∮Bdl cos90
∮ B.dl = BL + 0 + 0 + 0 ------1
-------2
from equation 1 and 2
μo × Inet = BL
Inet = n L I
μo × nLI = BL
BL = μo × nLI
B = μo nI
a
a
b
dcb
c d
x xxxxxx
Magnetic field B is perpendicular to sides bc and da
,hence these portions of the loop does not make any
contributions to the line integral as
B . dl = 0 for the side (b c ) and ( da )
Hence sum around the entire closed path reduces to Bl
Side cd lies at external points solenoid where B.dl = 0 as B=0
or negligibly small outside the solenoid.
a b
cd
∫B . dl= ∫B . dl+
ab
∫B . dl+
bc
∫B . dl+
cd
∫ B . dl
da
∫B . dl= μ0 Inet

10. •we have obtained this relation for infinitely long solenoids considering the field at external points of the
solenoid equal to zero.
B = 0
•However for real solenoids external field is relatively weak rather then equal to zero.
•Field at internal points of the solenoid does not depend on length and diameter of the solenoid and is
uniform over the cross-section of a solenoid.
The magnetic field inside a solenoid is
proportional to both the applied current and the
number of turns per unit length. There is no
dependence on the diameter of the solenoid, and
the field strength doesn't depend on the
position inside the solenoid, i.e., the field
inside is constant.
Magnetic field at the ends of a solenoid is half of
the magnetic field inside the solenoid.
B =
μo nI
𝟐

12. 3 Magnetic Field of a toroid
A solenoid bent into the form of a closed ring is called a toroidal solenoid.
• We will now apply Ampere circuital law to calculate magnetic field of a toroid .s
• A toroidal solenoid is a hollow circular ring with a large number of turns of a wire carrying current wound
around the ring.
• In this case Amperion loop would be a circle through point P and concentric inside the toroid
• By symmetry field will have equal magnitude at all points of this circle and this field is tangential to every
point in the circle.
• If there are total N number of turns ,net current crossing the area bounded by the circle is NI where I is the
current in the toroid.
• An endless solenoid in the form of a ring is called toroid magnetic field inside the toroid are circular
concentric with the centre of toroid .
• let I be the current , r be the mean radius and n be the number of turns per unit length and B be the
Magnetic field inside the toroid.
•Case-1: Magnetic field at a point in the empty space inside the toroid. We
will take an Amperian loop (loop 1).
By the Ampere’s circuital law:
We can see in the diagram above that current passing through the inside of the loop 1
is 0
∮B .dl = μo × current enclosed by closed path
∮ B1 . dl = μo × 0 = 0
∴ B1 = 0

13. . O
∮B .dl = μo × current enclosed by closed path
. P
. Q
.
Q
.
P
.
O
O is centre of toroid.
Loop 1 – of radius
Loop 2 – of radius
Loop 3 – of radius
Loop 1
Loop 2
Loop 3
r2
r1
r
r
r1
r2
I
dl
B
P
O
Q
B = 0
B = 0
r
B ≠ 0

14. • Case-2: Magnetic field at a point inside the toroid (between the turns). We will take another Amperian loop
(loop2) of radius r2.
•By the Ampere’s circuital law:
We can see in the diagram above that net current passing through the inside of the loop 2 is Ni, where N is
the total number of turns in the toroid
B2 ∮ dl = μo × Ni
B2 2πr2 = μo Ni
∴ B2 = μo Ni / (2πr2) = μo n i
Here n = number of turns per unit length of toroid = N / ( 2πr2 )
Note: The equation of magnetic field due to toroid is same as that of magnetic field due to solenoid.
•Case-3: Magnetic field at a point outside the toroid. We will take another Amperian loop (loop3) of
radius r3. By the Ampere’s circuital law:
We can see in the diagram above that net current passing through the inside of the loop 2 is 0 (Ni current
going out of the loop, and N i current entering the loop, so net current is O )
∮ B3.dl = μo×0 = 0
∴ B3 = 0
•Toroid are used in toroidal transformers, toroidal inductors etc.
∮B .dl = μo × current enclosed by closed path
∮B .dl = μo × current enclosed by closed path

15. • Thus we see that field B varies with r i.e. field B is not uniform over the cross-
section of the core because the path l = 2πr is longer at the outer side of the
section then at the inner side.
• Imagine a concentric circle through point P' outside the toroid.
• The net current passing through this circular disc is zero ,since the current NI
passes in and same current passes out. Thus using Ampere's circuital law, the
field B=0 outside the toroid.
NOTE:
The magnetic field exists only in the tubular area bound by the coil and it does not exist in the
area inside and outside the toroid.
i.e. B is zero at O and Q and non-zero at P.

16. Solenoid and toroid both work on the principle of electromagnetism and
both behave like an electromagnet when current is passed .
Solenoid
Cylindrical in shape.
Solenoid is considered as the
straight coil.
Magnetic field is created outside.
Has uniform magnetic field inside it.
Magnetic field due to solenoid is
B = μo nI
Toroid
Circular in shape.
Toroid is the bent solenoid which is having
ring or doughnut shape.
Magnetic field is created within.
It does not have a uniform magnetic field
inside it.
Magnetic field Outside :- magnetic field (B) =
0
Magnetic field Inside:-Magnetic field (B) = 0
Magnetic field Within the toroid:- Magnetic
field B = μo nI
In toroid due to the ring shape the magnetic field is much stronger than solenoid
at its center. Similarities between solenoid and toroid
1. Both works on the principle of electromagnetism.
2. When current is passed through them, they both act like an electromagnet.
3. Magnetic field due to the solenoid and within the toroid is the same. B = μo nI