Question

Obtain the formula for the magnetic field produced inside a very long current carrying solenoid wising Ampere's Circuital Law.

Answer 1

Answer:

this is ur answer hope it helps

Answer 2

The formula for the magnetic field is $\bold{\mu_0nI}$

Solution:

Number of turns in solenoid is n

Current passing through the solenoid is I

Magnetic field through the solenoid is B

Refer the image below for the direction of the current and position of solenoid.

By right hand rule, the magnetic field is parallel to the position of solenoid.

Magnetic field outside the solenoid:

Using Ampere's Circuital Law, we get,

$\bold{\oint \overrightarrow{B}\overrightarrow{dl}=\mu\times0}$

Since, the magnetic field is produced only within the solenoid not outside it.

$\bold{\therefore B = 0}$

Magnetic field inside the solenoid:

$\bold{\oint \overrightarrow{B}\overrightarrow{dl}=\int_{pq}\overrightarrow{B}\cdot \overrightarrow{dl}+\int_{qr}\overrightarrow{B}\cdot \overrightarrow{dl}+\int_{rs}\overrightarrow{B}\cdot \overrightarrow{dl}+\int_{sp}\overrightarrow{B}\cdot \overrightarrow{dl} \longrightarrow(1)}$

For path pq, magnetic and axis of the solenoid is along same direction.

$\bold{\int_{qr}\overrightarrow{B}\cdot \overrightarrow{dl}=\int_{sp}\overrightarrow{B}\cdot \overrightarrow{dl}=B.dl \cos 90^{\circ} = 0}$

For path rs, outside the solenoid, B = 0

$\bold{\int_{rs}\overrightarrow{B}\cdot \overrightarrow{dl} = 0}$

Now, the equation (1) becomes,

$\bold{\oint_{pqrs} \overrightarrow{B}\overrightarrow{dl}=\int_{pq}\overrightarrow{B}\cdot \overrightarrow{dl} = Bl}$

On using ampere law, we get,

$\bold{\oint \overrightarrow{B}\cdot \overrightarrow{dl} = \mu_0I}$

$\bold{Bl= \mu_0(nl\times I)}$

$\bold{\therefore B=\mu_0nI}$