Informative line

Self Inductance

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.

Self Inductance (L)

  • 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.

Factors affecting Inductance

(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.

Illustration Questions

Which of the following shape of inductor has more inductance when length of each coil and radius of each turns is same ?

A

B

C

D

×

More the number of turns, more will be the inductance.

Hence, option (B) is correct.

Which of the following shape of inductor has more inductance when length of each coil and radius of each turns is same ?

A image
B image
C image
D image

Option B is Correct

Average Induced EMF of Inductor when Current is Given in Terms of Time 

  • 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 

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

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

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

Unit of Inductance

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

Illustration Questions

The current in an inductor coil is given as \(I= 2\, t + 3 \) and the self inductance of the coil is \( L = 4 \,mH\). Find the average induced emf of the coil.

A 7 mV

B 9 mV

C 8 mV

D 10 mV

×

The average induced emf of coil is given as

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

where,

\(\mathcal{E}=\)Induced emf,

 L = Inductance, 

\(I\) = Current

Given:

\(I\) \(= 2\,t + 3\)\( L = 4\, mH\)

\(\mathcal{E}=4×10^{-3} \left [ \dfrac {d\;(2\,t+3)}{dt} \right]\)

\(\mathcal{E}=8×10^{-3}\;V=8\,mV\)

The current in an inductor coil is given as \(I= 2\, t + 3 \) and the self inductance of the coil is \( L = 4 \,mH\). Find the average induced emf of the coil.

A

7 mV

.

B

9 mV

C

8 mV

D

10 mV

Option C is Correct

Average Induced emf of Inductor

  • 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

Illustration Questions

The current in an inductor coil is changed from \(I_i=5\,A\) to \(I_f=10\,A\) in \(\Delta t=2\,sec.\) and the self inductance of the coil is given by \(L = 4\,mH\). Calculate the average induced emf of the coil.

A 100 V

B 50 V

C 10 V

D 0.01 V

×

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

Given : \(I_i=5\,A\)\(I_f=10\,A\)\(\Delta t=2\,sec.\),\( L = 4 \,mH\)

\(|\mathcal{E}_{avg}|=4×10^{-3}\dfrac {(10-5)}{2}\)

\(\mathcal{E}_{avg}=0.01\;Volt\)

The current in an inductor coil is changed from \(I_i=5\,A\) to \(I_f=10\,A\) in \(\Delta t=2\,sec.\) and the self inductance of the coil is given by \(L = 4\,mH\). Calculate the average induced emf of the coil.

A

100 V

.

B

50 V

C

10 V

D

0.01 V

Option D is Correct

Self Inductance of Solenoid

  • 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.

Illustration Questions

What will be the change in the self inductance of solenoid if the length of solenoid is increased and the number of turns are kept constant?

A Will increase

B Will decrease

C Will remain the same

D None of these

×

Since length of solenoid is inversely proportional to self inductance.

 \(L\propto \dfrac {1}{\ell}\)

So, if the length of solenoid is increased, the self inductance will decrease.

Hence option (B) is correct.

What will be the change in the self inductance of solenoid if the length of solenoid is increased and the number of turns are kept constant?

A

Will increase

.

B

Will decrease

C

Will remain the same

D

None of these

Option B is Correct

Induced E.M.F. when AC Flows in an Inductor Coil

  • 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

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

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

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

 

Illustration Questions

An alternating current \(I=3\) \(sin\;2\pi\; t\) is given to the inductor coil having self inductance \(L = 4\, mH\). Determine the maximum induced emf in the coil.

A 24\(\pi\)× 10–3 V

B 28 \(\pi\)× 10–3 V

C 26\(\pi\)× 10–3 V

D 20 \(\pi\)× 10–3 V

×

The induced emf is given as

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

where, \(L\) =Inductance,

\(I\) = Current

Given :   \(I=3sin\;2\pi\; t\) , \( L = 4 \,mH\)

\(\mathcal{E}=4×10^{-3}×\dfrac {d}{dt}\;(3\; sin\,2\pi t)\)

\(\mathcal{E}=4×10^{-3}×3×2\pi\; cos\,2\pi t\)

For maximum emf

\(cos\,2\pi t=1\)

\(\mathcal{E}=24\pi\; ×10^{-3} V\)

An alternating current \(I=3\) \(sin\;2\pi\; t\) is given to the inductor coil having self inductance \(L = 4\, mH\). Determine the maximum induced emf in the coil.

A

24\(\pi\)× 10–3 V

.

B

28 \(\pi\)× 10–3 V

C

26\(\pi\)× 10–3 V

D

20 \(\pi\)× 10–3 V

Option A is Correct

Illustration Questions

\(I-t\)  graph of an inductor coil having self inductance \(L = 3\,mH \) is given as shown in figure. Calculate the induced emf for the inductor from 0 to 3 sec.

A –5 mV

B –10 mV

C –15 mV

D –9 mV

×

Equation of \(I\)\(-\)\(t \) graph:

\(\Rightarrow\) \(I=\dfrac {(y_2-y_1)}{(x_2-x_1)}t\)

\(\Rightarrow\) \(I=\dfrac {(5-0)}{(3-0)}t\)

\(\Rightarrow\) \(I=\dfrac {5}{3}t\)

image

Induced emf is given as

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

where, \(L\) = Inductance

\(I\) = Current

image

Given: \(L = 3 \,mH\)\(I=\dfrac {5}{3}t\)

\(\mathcal{E}_{induced}=-3×10^{-3}\dfrac {d}{dt} \left (\dfrac {5}{3}t\right)\)

\(\mathcal{E}_{induced}=-3×10^{-3} \left (\dfrac {5}{3}\right)\)

\(\mathcal{E}_{induced}=\) \(–5\, mV\)

image

\(I-t\)  graph of an inductor coil having self inductance \(L = 3\,mH \) is given as shown in figure. Calculate the induced emf for the inductor from 0 to 3 sec.

image
A

–5 mV

.

B

–10 mV

C

–15 mV

D

–9 mV

Option A is Correct

Practice Now