prove that for
finite length current element ?H = I/2πr(sinα2 - sinα1)aφ
To prove that for a finite length current element, the magnetic field at a point in space can be represented by the equation H = I/2πr(sinα2 - sinα1)aφ, we can use Ampere's circuital law.
Ampere's circuital law states that the magnetic field around a closed loop is equal to the sum of the currents passing through the loop multiplied by the permeability of free space (μ₀).
To derive the expression for the magnetic field at a point due to a finite length current element, let's consider a small segment of the current element of length dl (located at a distance r from the point where we want to calculate the magnetic field) and take a closed loop around this segment.
The magnetic field contribution due to this small segment can be calculated using the Biot-Savart law. According to the Biot-Savart law, the magnetic field dB at a point due to a small segment of current is given by:
dB = (μ₀/4π) * (I dl x ẑ) / r²
Here, μ₀ is the permeability of free space, I dl is the current element, and ẑ is a unit vector pointing along the direction of current flow.
To determine the direction of dl and r in the cross product, let's consider a small angle dθ that represents the angle between the radial direction and the tangent to the current element. The vector dl is then directed tangentially along the current element. Since dl is tangential, the angle between dl and the radial direction is dθ + 90 degrees.
Using the small angle approximation, we can simplify the expression to:
dB = (μ₀/4π) * (I dl sin(dθ + 90°) / r²) aφ
Simplifying further, we get:
dB = (μ₀/4π) * (I dl sin(dθ) / r²) aφ
Integrating this equation over the whole length of the current element (from α1 to α2), we obtain:
∫ dB = ∫ (μ₀/4π) * (I dl sin(dθ) / r²) aφ
The left-hand side is the integral of the magnetic field over the closed loop, which is equal to the magnitude of the magnetic field multiplied by the circumference of the loop. Therefore, we have:
H * 2πr = (μ₀/4π) * ∫ (I dl sin(dθ) / r²) aφ
H * 2πr = (μ₀/4πr²) * ∫ I dl sin(dθ) aφ
Since sin(dθ) is a constant over the interval α1 to α2, we can factor it out of the integral:
H * 2πr = (μ₀/4πr²) * sin(dθ) * ∫ I dl aφ
The integral of I dl over the current element is simply the total current I:
H * 2πr = (μ₀/4πr²) * sin(dθ) * I aφ
sin(dθ) can be expressed in terms of angles α1 and α2 as sin(dθ) = sinα2 - sinα1. Substituting this expression, we get:
H * 2πr = (μ₀/4πr²) * (sinα2 - sinα1) * I aφ
Finally, rearranging the equation, we obtain the required expression for the magnetic field H:
H = I/2πr * (sinα2 - sinα1) aφ
Therefore, we have proved that for a finite length current element, the magnetic field at a point in space is given by the equation H = I/2πr * (sinα2 - sinα1) aφ.