Solenoid : a cylindrical coil of many tightly wound turns of insulated wire with diameter of the coil smaller than its length is called a solenoid.
Magnetic field due to current carrying solenoid :
Consider a very long solenoid having n turns per unit length of solenoid.Let current I be flowing through the solenoid.
The magnetic field inside the solenoid is almost uniform , strong and directed along.
The axis of the solenoid. The magnetic field outside a very long solenoid is very weak and can be neglected.
Step 1 : Let P be a point well within the solenoid. Consider any rectangular loop ABCD ( known as Amperian Loop ) passing through P.
Then
$\oint \overrightarrow{B}.\overrightarrow{dl}$ = Line integral of magnetic field across the loop ABCD,
$$\oint \overrightarrow{B}.\overrightarrow{dl}= \int_{A}^{B} \overrightarrow{B}.\overrightarrow{dl} + \int_{B}^{C} \overrightarrow{B}.\overrightarrow{dl} + \int_{C}^{D} \overrightarrow{B}.\overrightarrow{dl} + \int_{D}^{A} \overrightarrow{B}.\overrightarrow{dl}$$
$\overrightarrow{B}$ is perpendicular to paths BC and AD i.e. angle between $\overrightarrow{B}$ and $\overrightarrow{dl}$BN is 90° for these paths.
$\overrightarrow{B}.\overrightarrow{dl} + \int_{D}^{A} = \overrightarrow{B}.\overrightarrow{dl} + \int_{B}^{C} = \int B.dl.cos 90^{0} = 0$
Since path CD is outside the solenoid, where $\overrightarrow{B}$ is taken as zero , so
$\int_{C}^{D} \overrightarrow{B}.\overrightarrow{dl}= 0$
For path AB , the direction of $\overrightarrow{B}$ and $\overrightarrow{dl}$ is same $\theta = 0^{0}$
Hence equation (i) becomes
$\oint \overrightarrow{B}.\overrightarrow{dl} = = \int B.dl.cos 0^{0} = \int_{A}^{B} B.dl$
$\oint \overrightarrow{B}.\overrightarrow{dl} = B\int_{A}^{B}dl$
$\oint \overrightarrow{B}.\overrightarrow{dl} = Bl$
Step 2 : According to Ampere's circuital law
$\oint \overrightarrow{B}.\overrightarrow{dl} = \mu_{0} $× net current enclosed by loop ABCD
$\oint \overrightarrow{B}.\overrightarrow{dl} = \mu_{0} $× number of turns in the loop ABCD × I
=$ \mu_{0}nlI$
Step 3 : comparing equation (ii) and (iii) , we get
$Bl = =\mu_{0}nlI$
$B= =\mu_{0}nI$
Since n = \frac{N}{l}
Where N = total number of turns of solenoid
$B = \frac{\mu_{0}NI}{l}$
Tags:
Class-12-physics