$A$ long solenoid has $500$ turns. When a current of $2 \,A$ is passed through it, the resulting magnetic flux linked with each turn of the solenoid is $4 \times 10^{-3} \,Wb$. The self-inductance of the solenoid is: (in $\,H$)
- A$2.5$
- B$2.0$
- C$1.0$
- D$4.0$
Explore More
Similar Questions
The momentum in mechanics is expressed as $m \times v$. The analogous expression in electricity is
Easy
View Solution$A$ long coaxial cable carries current $I$ along its surface. The current flows down the surface of the inner cylinder of radius $a$ and back along the outer cylinder of radius $b$.
Difficult
View SolutionIn the given circuit,what happens when the switch $S$ is opened?
Easy
View Solution$A$ current $I$ produces a magnetic flux $\phi$ per turn in a coil of $n$ turns. The self-inductance of the coil is $L$. The relation between them is:
$A$ coil of wire of radius $r$ has $600$ turns and self-inductance of $108 \ mH$. The self-inductance of a coil with the same radius and $500$ turns is (in $mH$)
Vedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo