Question

A long solenoid has $1000turns$. When a current of $4A$ flows through it, the magnetic flux linked with each turn of the solenoid is $4×{10}^{-3}Wb$. What is the self-inductance of the solenoid?

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Solution

Step 1: GIven dataA long solenoid has $\left(N\right)=1000turns$Current $I=4A$Magnetic flux ${\varphi }_{\circ }=4×{10}^{-3}Wb$Step 2: Formula used$\varphi =LI$This formula belongs to the self-inductance of the solenoid.Here, $\varphi$=magnetic flux$L$= self-inductance$I$=currentStep 3: Calculating self-inductanceSo, we have flux linked with each turn ${\varphi }_{\circ }=4×{10}^{-3}Wb$Total flux linked =$1000×4×{10}^{-3}$Now, on the basis of the formula$\varphi =LI\phantom{\rule{0ex}{0ex}}L=\frac{\varphi }{I}$Now on putting the values$L=\frac{4×{10}^{-3}}{4}$So, $L=1H$Hence, the self-inductance of the solenoid is $L=1H$.

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