Question

The current through a 1 mh coil drops from 5 a to 3 a in 10-3s. the emf developed in the coil is:

Answer 1

The EMF developed in the coil is equal to 2 V.

Explanation:

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

A conducting coil resembles an inductor.

Given:

Inductance of the coil :  $L = 1\ mH= 1\times 10^{-3}\ H$

Initial current $I_1 = 5\ A$

Final current $I_2 = 3\ A$

Time = $t = 10^{-3} s$

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 developed due to an inductor coil is mathematically stated as

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

Rate of change of current :

$\frac{dI}{dt} = \frac{I_2 - I_1}{t} = \frac{3-5}{{10^{-3}}}\\\\\frac{dI}{dt} = - 2\times 10^3\ A/s$ .....................(2)

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

$E = - L\times \frac{dI}{dt}\\E = - 1\times 10^{-3}\times -2\times 10^3 = 2\ V$

Therefore, the emf developed in the coil is equal to 2 volts.