derivation of magnetic field due to finite current carrying solenoid
Answers
Answer: Formula 1 is the contribution to the magnetic field from the small segment of the coil of length dl shown in the lower portion of Figure 1. That piece of coil has current dI=NIdl/l flowing through it where N is the number of turns in the coil, I is the current per turn, and l is the length of the coil. From the hyperphysics site (or by integrating the Biot-Savart law), the axial field from this loop is:
Bz=μ04π(2πr)dIr(r2+z2)3/2=μ0r2dI2y3
where y=(r2+z2)1/2 is the distance from the measurement point to the rim of the loop and r is the coil radius. Substituting the formula for dI gives you the first equation.
The second equation is the integral of the above over the length l. The integral is first converted to an integral over angle θ via sinθ=r/y and sinθdl=ydθ:
B=∫μ0r2dI2ly3=μ0NI2l∫r2dly3=μ0NI2l∫θ2θ1sinθdθ
Explanation: