Question

in a circular conducting coil when current increases from 2 ampere to 18 ampere in 0.05 seconds The induced EMF is 20 volts the self inductance of the coil is​

Answer 1

The self inductance of the coil is 62.5 mH

Explanation:

Formula : $E= - L\times \frac{dI}{dt}$

A circular conducting coil resembles an inductor.

Given:

Induced emf :  $E = 20\ V$

Initial current : $I_1 = 2\ A$

Final current : $I_2 = 18\ A$

Time : $t = 0.05\ s$

An inductive element generates a back emf (self-induced emf) and it is defined as the product of self-inductance of the coil and the rate of change of current per unit time.

If

E = Induced emf

L = Self-inductance of the coil

$\frac{dI}{dt}$= Rate of change of current

The emf induced due to an inductor coil is mathematically stated as

$E = - L\times \frac{dI}{dt}$ ......................................(1)

Rate of change of current :

$\frac{dI}{dt} = \frac{I_2 - I_1}{t} = \frac{18-2}{{0.05}}}\\\\\frac{dI}{dt} = 320\ A/s$ .....................(2)

Substituting the value of emf (E) and rate of change of current from (2) in equation (1), we get

$E = - L\times \frac{dI}{dt}\\20 = -L\times 320 \\ L = - \frac{20}{320} = - 0.0625$

Inductance cannot be negative. Hence,

$L = 0.0625\ H = 62.5\ mH$

Therefore, the self inductance of the circular conducting coil is equal to 62.5 mH.

Answer 2

Answer:

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Explanation:

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