Amperes Law Formula
Ampère’s law is a fundamental principle in physics, particularly in the field of electromagnetism. It establishes a relationship between the integrated magnetic field around a closed loop and the electric current passing through that loop. In this article, we will explore the details and applications of Ampère’s circuital law.
What is Ampere’s Law Formula?
Ampère’s Law explains how magnetic fields are connected to the electric currents that generate them. This law specifies how to calculate the magnetic field resulting from a given electric current or vice versa.
Definition of Ampere’s Law
Ampere’s Law states that the magnetic field around a current-carrying conductor is directly proportional to the magnitude of the electric current, assuming the electric field remains constant over time. The constant of proportionality in this relationship is known as the permeability of free space.
The equation representing Ampère’s law, which is one of Maxwell’s equations, is expressed as:
∇×H= (∂D/∂t) +J
This equation describes the relationship between the curl of the magnetic field H, the time rate of change of the electric displacement field D, and the current density J.
What Is Ampere’s Circuital Law?
Ampère’s Circuital Law describes the connection between an integrated magnetic field around a closed loop and the electric current flowing through that loop. Derived from the Biot-Savart Law, Ampère’s Law offers an alternative method for determining the magnetic field generated by a specific current distribution. While the Biot-Savart Law focuses on calculating magnetic responses at the molecular and atomic levels, Ampère’s Law simplifies these calculations in many practical scenarios.
Definition of Ampere’s Circuital Law
Ampère’s Circuital Law states that the line integral of a magnetic field around a closed loop is equivalent to the sum of the currents passing through that loop.
The formula for Ampère’s Circuital Law is given by:
∮H⋅dl = μ0Ienc
Where:
- μ0 is the permeability of the medium.
- Ienc is the enclosed current.
Consider a conductor carrying a current I, which generates a magnetic field around the wire. Ampère’s law assumes that the closed loop consists of small differential elements of length dl.
On the left side of the equation, if an imaginary path encircles the wire and the magnetic field is summed at each point, this sum is numerically equal to the current enclosed by this path (Ienc ).
Ampere’s Law Formula
The total magnetic field density around the closed loop is the integral of the magnetic field times the differential length (B⋅dl).
This closed loop is also known as the Amperian Loop. The integral of the magnetic field along this loop is equal to the product of the net current passing through the loop and the permeability of the medium (μ0I). Thus:
∮B⋅dl=μ0I
This equation, derived by Maxwell, alternatively expresses that the magnetic field intensity (H) along an imaginary closed path equals the current enclosed by the path.
Therefore:
B⋅dl = μ0I
Bμ0⋅dl = I
H⋅dl = I
So:
H = Bμ0
Where,
- μ0 is the permeability constant with a value of 4π × 10-7 N/A2
- B is the magnetic field
- I is the flow of current passing through the closed loop
- L is the length of the loop
Ampère’s Circuital Law applies to a steady electric current that does not change with time.
Determining Magnetic Field by Ampere’s Law
The magnetic field at a distance r from a current-carrying wire can be determined using Ampère’s Law. For a wire conducting a current I, the magnetic field at a distance r is calculated using Ampère’s Law, and its direction is given by the Right Hand Thumb Rule.
To calculate the magnetic field around the wire, we draw an imaginary path (Amperian loop) at a distance r from the wire. According to the second Maxwell equation, the magnetic field integrated along this path equals the current enclosed by the wire, I.
In this scenario, the magnetic field remains constant at any point on the loop. The length of the path around the wire is 2πr. Thus, using Ampère’s Law:
∮B⋅dl=B⋅(2πr)=μ0I
Where:
- B is the magnetic field.
- dl is the differential length element.
- μ0 is the permeability of free space.
Therefore, the magnetic field B at a distance r from the wire is given by:
B= μ0I/2πr
This equation shows that the magnetic field around a long, straight conductor is inversely proportional to the distance r from the wire and directly proportional to the current I through the wire.
Applications of Ampere’s Law
Ampere’s Law has several applications:
- It is used to determine the magnetic field produced by a straight current-carrying wire.
- Ampere’s Law is applied to find the magnetic field within a toroidal coil.
- It helps in calculating the magnetic field inside a conductor that is carrying a current.
- Ampere’s Law is used to determine the force exerted between two parallel current-carrying conductors.
Limitations of Ampere’s Law
Ampere’s Law has several limitations:
- Ampere’s Law is primarily used to calculate the magnetic field generated by current-carrying conductors. It does not provide information about magnetic fields in regions without current.
- Ampere’s Law assumes that the electric field on the surface does not change with time. This means it is valid only in situations where the electric field is constant or quasi-static.
- The law is applicable only for steady (unchanging) currents. It is not suitable for analyzing time-varying currents or the dynamic fields present in electrodynamics.
Solved Examples on Ampere’s Law Formula
Example 1: Calculate the magnetic field at a distance of 0.1 m from a long, straight wire carrying a current of 5 A.
Solution:
Given:
Distance from the wire, r = 0.1 m
Current I = 5A
Permeability of free space, μ0 = 4π×10−7T⋅m/A
Using Ampere’s Law:
B= 𝜇0𝐼/2πr
Substitute the given values:
B= (4π×10−7 ×5)/2π×0.1
B=10−6
B=10×10−7
B=1×10−5T
Example 2: A toroid with a mean radius of 0.15 m has 500 turns and carries a current of 2 A. Calculate the magnetic field inside the toroid.
Solution:
Given:
Mean radius, r = 0.15 m
Number of turns, N = 500
Current, I=2A
Permeability of free space, μ0 = 4π×10−7 T⋅m/A
Using Ampere’s Law for a toroid:
B= 𝜇0𝑁𝐼/2πr
Substitute the given values:
B= (4π×10−7 ×500×2)/2π×0.15
B=1333.33×10−6
B=1.33×10−3 T
Example 3: A long solenoid has a density of 200 turns per centimeter and carries a current of 2.5 A. Determine the magnetic field at its center.
Solution:
Given data,
Current (I) = 2.5A.
Number of turns per cm (n) = 200
B=μ0nI
B=4π×10−7×200×10−2×2.5
B=6.28×10−2Wb/m2