Answer
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Hint: When the current is varied, the flux linked with the coil changes and an e.m.f is induced in the coil. It is given by
\[\varepsilon = - \dfrac{{d({N_{{\Phi _B}}})}}{{dt}}\]
Complete step by step solution:
It is the property of a coil by the virtue of which it possesses the growth or decay of induced current in a single coil. Mathematically, Magnetic flux changes in the coil is directly proportional to the induced current produced in the coil.
Now derivation of the expression for magnetic energy stored in the inductor of inductance L and current carrying I
When the current is varied, the flux linked with the coil changes and an e.m.f is induced in the coil. It is given by
\[\varepsilon = - \dfrac{{d({N_{{\Phi _B}}})}}{{dt}}\]
\[ = - L\dfrac{{dI}}{{dt}}\]
Now the rate of doing work is given by
\[\dfrac{{dW}}{{dt}} = \left| \varepsilon \right|I\]
Solving the given equation, we get,
\[ = LI\dfrac{{dI}}{{dt}}\]
Now integrating we get,
\[
W = \int {dW} \\
= \int\limits_0^I {LIdI} \\
\]
Thus, the total work done in establishing current from 0 to I is
\[\dfrac{1}{2}L{I^2}\] .
Additional information:
Self-Inductance—ϕ ∝ I where L is a proportionality constant which is called self-inductance. Hence, the coefficient of self-inductance is the ratio of flux changes in the coil to the induced current produced in the coil. Self-Inductance of a single coil is numerically equal to flux changes in the coil, provided the induced current provided in the coil is 1 ampere. Using Faraday Second law, Hence, self-Inductance of a single coil is the ratio of induced emf produced in the coil to the ratio of loss of induced current with respect to time.
Note: Self-inductance of a simple coil is numerically equal to induced emf produced in the coil provided ratio of induced current with respect to time. The self-induced e.m.f is also called back e.m.f as it opposes any change in current in the circuit. So, work needs to be done against back e.m.f in establishing Current. This work done is stored as magnetic potential energy.
\[\varepsilon = - \dfrac{{d({N_{{\Phi _B}}})}}{{dt}}\]
Complete step by step solution:
It is the property of a coil by the virtue of which it possesses the growth or decay of induced current in a single coil. Mathematically, Magnetic flux changes in the coil is directly proportional to the induced current produced in the coil.
Now derivation of the expression for magnetic energy stored in the inductor of inductance L and current carrying I
When the current is varied, the flux linked with the coil changes and an e.m.f is induced in the coil. It is given by
\[\varepsilon = - \dfrac{{d({N_{{\Phi _B}}})}}{{dt}}\]
\[ = - L\dfrac{{dI}}{{dt}}\]
Now the rate of doing work is given by
\[\dfrac{{dW}}{{dt}} = \left| \varepsilon \right|I\]
Solving the given equation, we get,
\[ = LI\dfrac{{dI}}{{dt}}\]
Now integrating we get,
\[
W = \int {dW} \\
= \int\limits_0^I {LIdI} \\
\]
Thus, the total work done in establishing current from 0 to I is
\[\dfrac{1}{2}L{I^2}\] .
Additional information:
Self-Inductance—ϕ ∝ I where L is a proportionality constant which is called self-inductance. Hence, the coefficient of self-inductance is the ratio of flux changes in the coil to the induced current produced in the coil. Self-Inductance of a single coil is numerically equal to flux changes in the coil, provided the induced current provided in the coil is 1 ampere. Using Faraday Second law, Hence, self-Inductance of a single coil is the ratio of induced emf produced in the coil to the ratio of loss of induced current with respect to time.
Note: Self-inductance of a simple coil is numerically equal to induced emf produced in the coil provided ratio of induced current with respect to time. The self-induced e.m.f is also called back e.m.f as it opposes any change in current in the circuit. So, work needs to be done against back e.m.f in establishing Current. This work done is stored as magnetic potential energy.
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