$A$ wire in the form of a tightly wound solenoid is connected to a $DC$ source and carries a current. If the coil is stretched so that there are gaps between successive elements of the spiral coil,will the current increase or decrease? Explain.
(A) When the solenoid is tightly wound,the magnetic field lines are confined within the solenoid,resulting in a high self-inductance $(L)$.
When the coil is stretched,the gaps between the turns increase,which causes the magnetic flux linkage per turn to decrease,leading to a decrease in the self-inductance $(L)$ of the solenoid.
According to the relation $V = L(di/dt) + iR$,for a fixed $DC$ voltage source,the steady-state current is determined by $I = V/R$.
However,during the stretching process,the change in self-inductance induces an electromotive force (emf) $\varepsilon = -L(di/dt) - i(dL/dt)$.
Since the self-inductance $L$ decreases $(dL/dt < 0)$,the induced emf opposes the change,effectively reducing the back-emf. Consequently,the current in the circuit increases.
When the coil is stretched,the gaps between the turns increase,which causes the magnetic flux linkage per turn to decrease,leading to a decrease in the self-inductance $(L)$ of the solenoid.
According to the relation $V = L(di/dt) + iR$,for a fixed $DC$ voltage source,the steady-state current is determined by $I = V/R$.
However,during the stretching process,the change in self-inductance induces an electromotive force (emf) $\varepsilon = -L(di/dt) - i(dL/dt)$.
Since the self-inductance $L$ decreases $(dL/dt < 0)$,the induced emf opposes the change,effectively reducing the back-emf. Consequently,the current in the circuit increases.
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Similar Questions
Current in a circuit falls from $5.0\, A$ to $0\, A$ in $0.1\, s$. If an average $emf$ of $200\, V$ is induced,estimate the self-inductance of the circuit in $H$.
Easy
View SolutionThe length of a wire required to make a solenoid of length $l$ and self-induction $L$ is
The self-inductance of a long solenoid of cross-sectional area $A$,length $l$ and $n$ turns per unit length is given by
This question has Statement $1$ and Statement $2.$ Of the four choices given after the Statements,choose the one that best describes the two Statements.
Statement $1 :$ Self inductance of a long solenoid of length $L,$ total number of turns $N$ and radius $r$ is less than $\frac{{\pi {\mu _0}{N^2}{r^2}}}{L}$
Statement $2:$ The magnetic induction in the solenoid in Statement $1$ carrying current $I$ is $\frac{{{\mu _0}NI}}{L}$ in the middle of the solenoid but becomes less as we move towards its ends.
Statement $1 :$ Self inductance of a long solenoid of length $L,$ total number of turns $N$ and radius $r$ is less than $\frac{{\pi {\mu _0}{N^2}{r^2}}}{L}$
Statement $2:$ The magnetic induction in the solenoid in Statement $1$ carrying current $I$ is $\frac{{{\mu _0}NI}}{L}$ in the middle of the solenoid but becomes less as we move towards its ends.
MediumAIEEE 2012
View SolutionThe 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 $\left[\mu_0 = 4 \pi \times 10^{-7} \ SI \ unit\right]$.
DifficultMHT CET 2021
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