When the number of turns in a coil is doubled | vedclass

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

Question

(A) The self-inductance $L$ of a solenoid is given by the formula $L = \frac{\mu_0 N^2 A}{l}$,where $N$ is the number of turns,$A$ is the cross-sectio

Vedclass · Accepted Answer

Four times

Vedclass · Answer

(A) The self-inductance $L$ of a solenoid is given by the formula $L = \frac{\mu_0 N^2 A}{l}$,where $N$ is the number of turns,$A$ is the cross-sectional area,and $l$ is the length of the coil.Since $A$ and $l$ remain constant,we have $L \propto N^2$.Given that the number of turns is doubled,$N_2 = 2N_1$.Therefore,the new self-inductance $L_2$ is given by $L_2 = L_1 \left( \frac{N_2}{N_1} \right)^2 = L_1 \left( \frac{2N_1}{N_1} \right)^2 = 4L_1$.Thus,the self-inductance becomes four times the original value.

When the number of turns in a coil is doubled without any change in the length of the coil,its self-inductance becomes

Similar Questions

The current through a coil of self-inductance $L = 2 \ mH$ is given by $I = t^2 e^{-t}$ at time $t$. How long will it take for the induced electromotive force $(emf)$ to become zero (in $s$)?

Current in a circuit falls from $5.0 \ A$ to $0.0 \ A$ in $0.1 \ s$. If an average emf of $100 \ V$ is induced,then the value of self-inductance of the circuit is . . . . . . . (in $H$)

The current passing through a coil of $120$ turns and inductance $40 \text{ mH}$ is $30 \text{ mA}$. The magnetic flux linked with the coil is:

$A$ graph of magnetic flux $(\phi)$ versus current $(I)$ is plotted for four inductors $A, B, C, D$. Which inductor has the largest value of self-inductance?

Consider a solenoid carrying current supplied by a $DC$ source with a constant $emf$ containing an iron core inside it. When the core is pulled out of the solenoid,the change in current will:

Vedclass Test Series

Exam Paper Generator

Online Exam Module