If the current of $I$ gives rise to a magnetic flux $\phi$ through a coil having $N$ turns,then the magnetic energy stored in the medium surrounding the coil is
- A$\frac{N \phi I}{4}$
- B$\frac{N \phi I}{2}$
- C$\frac{NI}{2 \phi}$
- D$\frac{N \phi}{2 I}$
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Similar Questions
Assertion : If the current in a solenoid is reversed in direction while keeping the same magnitude,the magnetic field energy stored in the solenoid remains unchanged.
Reason : Magnetic field energy density is proportional to the magnetic field.
Reason : Magnetic field energy density is proportional to the magnetic field.
If the current in an $LR$ circuit is reduced to half,what will be the energy stored in it?
$A$ $100 mH$ coil carries a current of $1 A$. Energy stored in the form of magnetic field is (in $J$)
$(a)$ Obtain the expression for the magnetic energy stored in a solenoid in terms of magnetic field $B$,area $A$,and length $l$ of the solenoid.
$(b)$ How does this magnetic energy compare with the electrostatic energy stored in a capacitor?
$(b)$ How does this magnetic energy compare with the electrostatic energy stored in a capacitor?
Medium
View SolutionThe magnetic energy stored in an inductor of inductance $4\,\mu H$ carrying a current of $2\,A$ is:
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