define self inductance of a coil . show that magnetic energy required to build up the current in a coil of self inductance is given by 1/2LII

Open in App

Solution

$Answer When a current is established in a conductor, a magnetic field is produced in its vicinity . We can visualize this field in terms of magnetic flux . If steady current flows the number of lines of force at a given place would remain the same . But if the current changes with time, the flux associated with the loop changes and hence an e . m . f is induced in the loop . This phenomenon is called as self induction . Self - induction is the property of a coil, which enables the induction of an EMF in it when the current in the coil changes . L \frac{μ_{0} N^{2} A}{I} Consider a coil carrying a current I having N turns and lets the magnetic flux f be linked with the coil . If the current in the coil is changed, the flux link also changes . Thus, according to Faraday' s law of electromagnetic induction, induces an EMF on to itself . This EMF is called self - induced EMF and this phenomenon is called self - induction . It is found that the flux linkage is proportional to the current through it . E = - \frac{d ϕ}{d t} E = - L \frac{d I}{d t} Energy stored in inductor If' i' amperes of current is established in a coil in time' dt', then' dw' amount of work done if' e' stands for the induced e . m . f . d w = - e I d t d w = L \frac{d I}{d t} I d t = L I d I the total work done to build up current from 0 to I w = \int dw = \int_{0}^{I} LIdI = \frac{1}{2} {LI}^{2}$

Suggest Corrections

Similar questions

\frac{1}{2} L I^{2}

L

I

4 A

L = 200 m H

Join BYJU'S Learning Program