# sir,                       I cant able to understand this formula for field along the axis of a solenoid                                                           B=   NI(cos2-cos1)/2 also please derive for me the conditions when the solenoid is long and at the end s of the solenoid.

ROSHAN MUJEEB
3 years ago
A finite solenoid is a solenoid with finite length. Continuous means that the solenoid is not formed by discrete coils but by a sheet of conductive material. We assume the current is uniformly distributed on the surface of the solenoid, with a surfacecurrent densityK; incylindrical coordinates:

{\displaystyle {\vec {K}}={\frac {I}{l}}{\hat {\phi }}.} [{\displaystyle {\vec {K}}={\frac {I}{l}}{\hat {\phi }}.}]

The magnetic field can be found using thevector potential, which for a finite solenoid with radiusRand lengthlin cylindrical coordinates{\displaystyle (\rho ,\phi ,z)} [(\rho, \phi, z)] is[5]

{\displaystyle A_{\phi }={\frac {\mu _{0}I}{4\pi }}{\frac {1}{l}}{\sqrt {\frac {R}{\rho }}}\left[\zeta k\left({\frac {k^{2}+h^{2}-h^{2}k^{2}}{h^{2}k^{2}}}K(k^{2})-{\frac {1}{k^{2}}}E(k^{2})+{\frac {h^{2}-1}{h^{2}}}\Pi (h^{2},k^{2})\right)\right]_{\zeta _{-}}^{\zeta _{+}},} [{\displaystyle A_{\phi }={\frac {\mu _{0}I}{4\pi }}{\frac {1}{l}}{\sqrt {\frac {R}{\rho }}}\left[\zeta k\left({\frac {k^{2}+h^{2}-h^{2}k^{2}}{h^{2}k^{2}}}K(k^{2})-{\frac {1}{k^{2}}}E(k^{2})+{\frac {h^{2}-1}{h^{2}}}\Pi (h^{2},k^{2})\right)\right]_{\zeta _{-}}^{\zeta _{+}},}]

where

{\displaystyle \zeta _{\pm }=z\pm {\frac {l}{2}},} [{\displaystyle \zeta _{\pm }=z\pm {\frac {l}{2}},}]
{\displaystyle h^{2}={\frac {4R\rho }{(R+\rho )^{2}}},} [{\displaystyle h^{2}={\frac {4R\rho }{(R+\rho )^{2}}},}]
{\displaystyle k^{2}={\frac {4R\rho }{(R+\rho )^{2}+\zeta ^{2}}},} [{\displaystyle k^{2}={\frac {4R\rho }{(R+\rho )^{2}+\zeta ^{2}}},}]
{\displaystyle K(m)=\int _{0}^{\pi /2}{\frac {1}{\sqrt {1-m\sin ^{2}\theta }}}d\theta ,} [K(m)=\int_0^{\pi/2}{\frac{1}{\sqrt{1-m \sin^2 \theta }}} d\theta,]
{\displaystyle E(m)=\int _{0}^{\pi /2}{\sqrt {1-m\sin ^{2}\theta }}d\theta ,} [E(m)=\int_0^{\pi/2}{\sqrt{1-m \sin^2 \theta} } d\theta,]
{\displaystyle \Pi (n,m)=\int _{0}^{\pi /2}{\frac {1}{(1-n\sin ^{2}\theta ){\sqrt {1-m\sin ^{2}\theta }}}}d\theta .} [\Pi(n,m)=\int_0^{\pi/2}{\frac{1}{(1-n \sin^2 \theta)\sqrt{1-m \sin^2 \theta }}} d\theta.]

Here,{\displaystyle K(m)} [K(m)] ,{\displaystyle E(m)} [E(m)] , and{\displaystyle \Pi (n,m)} [\Pi(n,m)] are completeelliptic integralsof the first, second, and third kind.

Using

{\displaystyle {\vec {B}}=\nabla \times {\vec {A}},} [\vec{B} = \nabla \times \vec{A},]

the magnetic flux density is obtained as[6][7][8]

{\displaystyle B_{\rho }={\frac {\mu _{0}I}{4\pi }}{\frac {2}{l}}{\sqrt {\frac {R}{\rho }}}\left[{\frac {k^{2}-2}{k}}K(k^{2})+{\frac {2}{k}}E(k^{2})\right]_{\zeta _{-}}^{\zeta _{+}},} [{\displaystyle B_{\rho }={\frac {\mu _{0}I}{4\pi }}{\frac {2}{l}}{\sqrt {\frac {R}{\rho }}}\left[{\frac {k^{2}-2}{k}}K(k^{2})+{\frac {2}{k}}E(k^{2})\right]_{\zeta _{-}}^{\zeta _{+}},}]
{\displaystyle B_{z}={\frac {\mu _{0}I}{4\pi }}{\frac {1}{l}}{\frac {1}{\sqrt {R\rho }}}\left[\zeta k\left(K(k^{2})+{\frac {R-\rho }{R+\rho }}\Pi (h^{2},k^{2})\right)\right]_{\zeta _{-}}^{\zeta _{+}}.} [{\displaystyle B_{z}={\frac {\mu _{0}I}{4\pi }}{\frac {1}{l}}{\frac {1}{\sqrt {R\rho }}}\left[\zeta k\left(K(k^{2})+{\frac {R-\rho }{R+\rho }}\Pi (h^{2},k^{2})\right)\right]_{\zeta _{-}}^{\zeta _{+}}.}]

On the symmetry axis, the radial component vanishes, and the axial field component is

{\displaystyle B_{z}={\frac {\mu _{0}NI}{2}}{\Biggl (}{\frac {l/2-z}{l{\sqrt {R^{2}+(l/2-z)^{2}}}}}+{\frac {l/2+z}{l{\sqrt {R^{2}+(l/2+z)^{2}}}}}{\Biggr )}} [{\displaystyle B_{z}={\frac {\mu _{0}NI}{2}}{\Biggl (}{\frac {l/2-z}{l{\sqrt {R^{2}+(l/2-z)^{2}}}}}+{\frac {l/2+z}{l{\sqrt {R^{2}+(l/2+z)^{2}}}}}{\Biggr )}}] .

Inside the solenoid, far away from the ends ({\displaystyle |z|\ll l/2-R} [{\displaystyle |z|\ll l/2-R}] ), this tends towards the constant value{\displaystyle B=\mu _{0}NI/l} [{\displaystyle B=\mu _{0}NI/l}] .