
Define self-inductance of a coil. Show that magnetic energy required to build up the current \[I\] in a coil of self-inductance \[L\] is given by \[\dfrac{1}{2}L{{I}^{2}}\].
Answer
565.2k+ views
Hint: Inductance is the tendency of an electrical device to oppose a change in the electric current flowing through it. So the definition of self-inductance can now be framed accordingly. Magnetic energy is the energy stored in the magnetic field. With increase in current the inductance increases.
Complete step by step solution:
Definition of self-inductance: The induction of a voltage in a current-carrying wire or coil when the current in the wire itself is changing is known as self-inductance. In the case of self-inductance, the magnetic field created by a changing current in the circuit is responsible for inducing a voltage in the same circuit.
Self-inductance is numerically equal to the magnetic flux linked with the coil when current flows through it. Mathematically,
\[\Phi =Li\]
Where
\[\Phi \] is the flux linked with the coil
\[L\] is the constant of proportionality and it is called the self-inductance
\[i\] is the current flowing in the coil
Magnetic Energy stored in an inductor:
Consider a source having emf connected to an inductor,
With increase in current, the opposing induced emf increases and it is given as:
\[e=-L\dfrac{di}{dt}\]
Where \[e\] is the emf of the coil and current \[di\] flowing through it in the \[dt\] time period.
The negative sign indicated that inductance is opposing the change in electrical current.
When the source send current \[i\] through the inductor in small time \[dt\] the work done \[W\] by the source will be given as
\[dW=\int\limits_{0}^{I}{Lidi}\]
Hence the total work done will be given as:
\[W=\int\limits_{0}^{I}{Lidi}\]
\[\Rightarrow W=L\int\limits_{0}^{I}{idi}\]
\[\Rightarrow W=\dfrac{1}{2}L{{I}^{2}}\]
This work done is stored in the inductor in the form of Energy
Therefore the total energy stored in the magnetic inductor is given as \[U=\dfrac{1}{2}L{{I}^{2}}\].
Note: Do remember that the work done by the current is stored in the inductor in the form
of magnetic energy. Inductor opposes the change in the electric current. The change in the electric current is responsible for the self-inductance in the inductor. Also remember that the limits of current are taken from zero to some maximum value \[I\].
Complete step by step solution:
Definition of self-inductance: The induction of a voltage in a current-carrying wire or coil when the current in the wire itself is changing is known as self-inductance. In the case of self-inductance, the magnetic field created by a changing current in the circuit is responsible for inducing a voltage in the same circuit.
Self-inductance is numerically equal to the magnetic flux linked with the coil when current flows through it. Mathematically,
\[\Phi =Li\]
Where
\[\Phi \] is the flux linked with the coil
\[L\] is the constant of proportionality and it is called the self-inductance
\[i\] is the current flowing in the coil
Magnetic Energy stored in an inductor:
Consider a source having emf connected to an inductor,
With increase in current, the opposing induced emf increases and it is given as:
\[e=-L\dfrac{di}{dt}\]
Where \[e\] is the emf of the coil and current \[di\] flowing through it in the \[dt\] time period.
The negative sign indicated that inductance is opposing the change in electrical current.
When the source send current \[i\] through the inductor in small time \[dt\] the work done \[W\] by the source will be given as
\[dW=\int\limits_{0}^{I}{Lidi}\]
Hence the total work done will be given as:
\[W=\int\limits_{0}^{I}{Lidi}\]
\[\Rightarrow W=L\int\limits_{0}^{I}{idi}\]
\[\Rightarrow W=\dfrac{1}{2}L{{I}^{2}}\]
This work done is stored in the inductor in the form of Energy
Therefore the total energy stored in the magnetic inductor is given as \[U=\dfrac{1}{2}L{{I}^{2}}\].
Note: Do remember that the work done by the current is stored in the inductor in the form
of magnetic energy. Inductor opposes the change in the electric current. The change in the electric current is responsible for the self-inductance in the inductor. Also remember that the limits of current are taken from zero to some maximum value \[I\].
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