When current in a coil changes from 2 A to 5 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The formula for the induced emf in a coil due to self-inductance is given by $\varepsilon = -L \frac{di}{dt}$.Taking the magnitude,we have $|\vare

Vedclass · Accepted Answer

$4 \ mH$

Vedclass · Answer

(B) The formula for the induced emf in a coil due to self-inductance is given by $\varepsilon = -L \frac{di}{dt}$.Taking the magnitude,we have $|\varepsilon| = L \frac{|\Delta i|}{\Delta t}$.Given:Initial current $i_1 = 2 \ A$Final current $i_2 = 5 \ A$Change in current $\Delta i = i_2 - i_1 = 5 \ A - 2 \ A = 3 \ A$.Time interval $\Delta t = 0.3 \ s$.Induced emf $\varepsilon = 40 \ mV = 40 	imes 10^{-3} \ V$.Substituting the values into the formula:$40 	imes 10^{-3} = L 	imes \frac{3}{0.3}$.$40 	imes 10^{-3} = L 	imes 10$.$L = \frac{40 	imes 10^{-3}}{10} = 4 	imes 10^{-3} \ H$.$L = 4 \ mH$.

When current in a coil changes from $2 \ A$ to $5 \ A$ in a time of $0.3 \ s$,if the emf induced in the coil is $40 \ mV$,then the self inductance of the coil is

Similar Questions

The self-inductance of a straight conductor is

The current flowing through an inductance coil of self-inductance $6 \text{ mH}$ at different time instants is as shown. The emf induced between $t=20 \text{ s}$ and $t=40 \text{ s}$ is nearly

Consider a toroid with a rectangular cross-section,of inner radius $a$,outer radius $b$,and height $h$,carrying $n$ number of turns. Then the self-inductance of the toroidal coil when current $I$ is passing through the toroid is:

The self-inductance of a solenoid of length $31.4 \ cm$,area of cross-section $10^{-3} \ m^2$ having a total number of turns $500$ will be nearly $\left[\mu_0 = 4 \pi \times 10^{-7} \ SI \ unit\right]$.

The current through a coil varies according to the following graph. Plot the graph for induced $emf$ versus time for the coil.

Vedclass Test Series

Exam Paper Generator

Online Exam Module