Question

Consider a rectangular loop ABCD. For this loop $\frac{\oint }{ABCD}B,\underset{d\ell }{\overset{}{\to }}={\mu }_o{I}_{vps}$

Now

$\frac{\oint }{ABCD}\overline{B}\overline{d\ell }=\oint \overline{B}\overline{d\ell }+\frac{\oint }{ac}\overline{B.}\overline{d\ell }+\frac{\oint }{CD}\overline{B}\overline{d\ell }+\frac{\oint }{DA}\overline{B.}\overline{d\ell }=B\times a$

This is because $\frac{\oint }{AB}\overline{B}\overline{d\ell }=\frac{\oint }{CD}\overline{B}\overline{d\ell }=0,\overline{B}\perp \overline{d\ell }$

$And,\frac{\oint }{DA}\overline{B}\overline{d\ell }=0$

$(\therefore \overline{B}outside$ the solenoid in negligible Now,

$I_{mg}=(n\times a)\times is\Rightarrow B\times a=\mu \_o(n\times a\times i)$

$\Rightarrow B={\mu }_{o}ni$

Finite length solenoid :

Answer 1

Magnetic Field on the Axis of a Solenoid Having $N$ Turns Per Unit Length and Carrying a Current $I$

The field at a point on the axis of a solenoid can be obtained by superposition of fields due to a large number of identical coils all having their center on the axis of solenoid.

The magnetic field due this coil

$dB=\frac{{\mu }_{0}NI{R}^2}{2{(R^2+x^2)}^{3/2}}$

$dN=ndx,x=R\tan \varphi \text{and}dx=R{\sec }^2\varphi d\varphi$

$dB=\frac{{\mu }_{0}ndx\times I{R}^2}{2{(R^2+R^2{\tan }^2\varphi )}^{3/2}}$

$B=\int dB=\frac{{\mu }_0}{2}nI{\int }_{\alpha }^{\beta }\cos \varphi \text{d}\varphi$

$B=\frac{{\mu }_{0}nI}{2}[\sin \alpha +\sin \beta ]$

For a point inside a long solenoid, $\alpha =\beta ={90}^{{}^{{}^{\circ }}}$ therefore, $B={\mu }_{0}nI.$

At one end of a long solenoid, $\alpha =0^{{}^{{}^{\circ }}},b={90}^o$ therefore, $B={\mu }_{0}nI/2$