$A$ current of $2\,A$ is increasing at a rate of $4\,A/s$ through a coil of inductance $2\,H$. The energy stored in the inductor per unit time is ... $J/s$.
- A$2$
- B$1$
- C$16$
- D$4$
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The expression for the magnetic energy per unit volume stored in a solenoid in terms of magnetic field is . . . . . . .
What is the energy stored in the inductor of the given circuit?
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View SolutionEnergy stored in a coil of self-inductance $40 \, mH$ carrying a steady current of $2 \, A$ is.....$J$
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View SolutionIf a current of $4 \ A$ produces a magnetic flux of $3 \times 10^{-3} \ Wb$ through a coil of $400$ turns,the energy stored in the coil will be: (in $J$)
The energy stored in a $50 \, mH$ inductor carrying a current of $4 \, A$ will be $J$.
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