Learn the change in the self inductance of solenoid if the length of solenoid is increased and the number of turns are kept constant. Practice formula of Self Inductance equation.

- A current in a closed conducting loop produces a magnetic field around it.
- This magnetic field has its own flux through area bounded by the loop.
- Thus if the current changes with time, the flux also changes.

\(\phi_m\propto I\)

\(\phi_m=L\; I\)

where L = Self inductance

- As the current increases in the loop, the magnetic field increases and thus the flux also increases.

**NOTE:- **

- The induced current does not oppose the current through the inductor, it opposes the change in current in the solenoid.
- With increase or decrease in current, an opposite current is induced to oppose the change.
- This is known as inertia of current which opposes the change in current.

(1) More number of turns results in greater inductance.

**REASON:** More turns of wire generate greater amount of magnetic field for a given current. Thus, more flux will be generated which leads to more inductance.

\(\phi=L I\)

(2) Larger the area of coil, greater will be inductance.Given the number of turns are constant.

**REASON:** As area of coil increases, flux increases. Thus in turn inductance increases.

- A current in a closed conducting loop produces a magnetic field around it.
- This magnetic field has its own flux through area bounded by the loop.
- Thus if the current changes with time, the flux also changes.

\(\phi_m\propto I\)

\(\phi_m=L\; I\)

where L = Self inductance

- As the current increases in the loop, the magnetic field increases and thus the flux also increases.
- The induced emf due to change of current in the coil is given as

\(\mathcal{E}=-\dfrac {d\phi}{dt}\)

\(\mathcal{E}=-\dfrac {d}{dt}(LI)\)

\(\mathcal{E}=-L\dfrac {dI}{dt}\)

Units of L = Wb/Amp. or Henry (H)

- Average induced emf is given as

\(|\mathcal{E}_{avg}|=L\dfrac {\Delta I}{\Delta t}\)

\(|\mathcal{E}_{avg}|=L\dfrac {I_f-I_i}{\Delta t}\),

where, \(L \)= Self inductance,

\(I_f=\) Final value of current,

\(I_i=\)Initial value of current

\(\Delta t=\) Time interval

- Consider a long solenoid of radius "r" having N turns and total length \(\ell\). Current \(I\) is passing through it.
- The magnetic field produced inside the solenoid is \(B=\dfrac {\mu _0NI}{\ell}\)
- The flux through each turn is \(\phi=BA\)

\(\phi=\dfrac {\mu_0NI}{\ell}\cdot\pi r^2\)

- Emf induced in each turn is given as \(\mathcal{E}=-\dfrac {d\phi}{dt}\)

\(\mathcal{E}=-\dfrac {\mu_0N\;\pi r^2}{\ell}×\dfrac {dI}{dt}\)...(1)

We know, \(\mathcal{E}=-L\dfrac {dI}{dt}\) ...(2)

Comparing equation (1) and (2)

\(L_{each}=\dfrac {\mu_0N}{\ell}\;\pi r^2\) for each turn

- When the length of solenoid has N turns then,

\(L=N \left (\dfrac {\mu_0N}{\ell}\;\pi r^2\right)=\dfrac {\mu_0N^2}{\ell}\;\pi r^2\)

**NOTE:** Self inductance depends only on dimensions and geometrical factors.

A Will increase

B Will decrease

C Will remain the same

D None of these

- Consider a current \(I\) = \(I_0\;sin\,\omega t\) is flowing in an inductor coil having self inductance L.
- The induced emf is given as

\(\mathcal{E}=L\dfrac {dI}{dt}\)

\(\mathcal{E}=L\dfrac {d}{dt}(I_0\;sin\,\omega t)\)

\(\mathcal{E}=L\;I_0\,\omega\;cos\,\omega t\)

A 24\(\pi\)× 10–3 V

B 28 \(\pi\)× 10–3 V

C 26\(\pi\)× 10–3 V

D 20 \(\pi\)× 10–3 V