Derive equation of magnetic field inside a long straight solenoid.

Explanation

1. Figure shows the sectional view of a long solenoid. At various turns of the solenoid, current comes out of the plane of paper at points marked and enters the plane of paper at points marked .

2. To determine the magnetic field `vecB` at any inside point, consider a rectangular closed path abcd as the Amperean loop.
3. According to Ampere.s circuital law,
`ointvecB*vec(dl)=int_(a)^(b)vecB*vec(dl)+int_(b)^(c)vecB*vec(dl)+int_(c)^(d)vecB*vec(dl)+int_(d)^(a)vecB*vec(dl)" "...(1)`
4. cd part is outside the solenoid. Outside magnetic field `|vecB|=O` so that,
`int_(c)^(d)vecB*vec(dl)=O`
5. d -a and b - c parts are perpendicular to `vecB` so,
`int_(b)^(c)vecB*vec(dl)=int_(d)^(a)vecB*vec(dl)=O`
6. So equation (1) will be like this,
`ointvecB*vec(dl)=int_(a)^(b)vecB*vec(dl)=int_(a)^(b)BdlcosO=Bint_(a)^(b)dl`
`thereforeointvecB*vec(dl)=B(h)" "...(2)` (Where h = length of part ab)
7. Suppose, number of turn per unit length be n. So total number nh in h length.
8. I be current to each turn, so nhI current for nh turns current enclosed by loop `I_(e)=I(nh)" "...(3)`
9. According to Ampere.s circuital law,
`ointvecB*vec(dl)=mu_(0)I_(e)" "...(4)`
`therefore` Substitute equation (2) and (3) in equation (4),
`B(h)=mu_(0)Inh`
`thereforeB=mu_(0)nI" "...(5)`
10. The direction of the field is given by the right hand rule.
11. The solenoid is commonly used to obtain uniform magnetic field.