Show that the magnetic field at any point on the axis of the solenoid having n turns per unit length is B =`(1)/(2) mu_(0) n I (cos theta_(1) - cos theta_(2))`.

Explanation

A solenoid is a cylindrical coil having number of circular turns. Consider a solenoid having radius R consists of n number of turns per unit length .
Let .P. be the point at a distance .x. from the origin of the solenoid . the current carrying element dx at a distance x from origin and the distance r from point .P. .
r = `sqrt(R^(2) + (x_(0) - x)^(2))`
The magnetic field due to current carrying circular coil at any axis is
dB = `(mu_(0) I R^(2))/(2 r^(3) ) xx `N
Where N = ndx, then
dB = `(mu_(0))/(2) (n I R^(2) dx)/(r^(2))`
sin `theta = (R)/(r)`
r = R `xx (l)/(sin theta)` = R cosec `theta`
`tan theta = (R)/(x_(0) -x)`
`x_(0) - x = R xx (l)/(tan theta) = R cot theta`
`(dx)/(d theta) R "cosec"^(2) theta rArr " dx " = R " cosec"^(2) theta d theta " "`...(3)
From above three equations, we get
dB = `(mu_(0))/(2) (n I R^(2) (R "cosec"^(2) theta d theta))/(R^(3) "cosec"^(3)theta) `
dB = `(mu_(0))/(2) n I sin theta d theta`
Now total magnetic field can be obtained by integrating from `theta_(1) to theta_(2)`, we get
`B = (mu_(0) n I )/(2) int_(theta_(1))^(theta_(2)) sin theta d theta = (mu_(0) n I)/( 2) [ - cos theta]_(theta 1)^(theta 2)`
` B = - (mu_(0)n I)/(2 ) [ cos theta_(2) - cos theta_(1) ]`
B ` = (mu_(0) n I )/(2) [ cos theta_(1) - cos theta_(2) ] `