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I think you're relating an Inductor coil with Magnetic Lorentz force.

Specifically, Lorentz force is experienced by a charged particle moving in an uniform magnetic field. $F=q(\vec{v}\times\vec{B})$. Let's see Inductor first...

Please keep in mind that an Inductor always has a self-inductance $L$ and some resistance $R$ (of the material it's made of - except for an ideal one..!) associated with it. Hence, Some work has to be done by external agencies in establishing current. This work done is stored as electromagnetic potential energy in an inductor. In case of capacitor, it's electrostatic..!

Induced emf, $e=-L\frac{dI}{dt}$ (negative sign indicates the opposing nature, consequence of Lenz law)

The small amount of work done over a small time $dt$ is $$dw=e.I.dt$$ $$\implies dw=-LI.dI$$ (As @Alfred said, PowerPower $P=VI$ is used here, because Power is the rate of doing work as we know...)

Thus, the total work done is establishing a steady current (say $I_o$) is $$W=-L\int_0^{I_0}I.dI=-\frac{1}2LI_o^2$$

Negative sign shows the opposing nature of the emf (asame consequence of Lenz law) Thus, it has somewhat the same expression as that of a capacitor $\frac{1}2CV^2$. The small time $dt$ is taken into account because - Whenever you pass current through an inductor, steady current would be established only after some period of time. The current would be established in an increasing order. When the power supply is provided, the induced emf opposes the growth of current and when the power supply is cut-off, the emf now opposes the decay of current. (i.e. It goes in a decreasing order)

We could conclude thatHence, the work done by these agencies is referred to as the energy stored in an inductor. Now, do you still relate both of them which are completely non-correlated..?

I think you're relating an Inductor coil with Magnetic Lorentz force.

Specifically, Lorentz force is experienced by a charged particle moving in an uniform magnetic field. $F=q(\vec{v}\times\vec{B})$. Let's see Inductor first...

Please keep in mind that an Inductor always has a self-inductance $L$ and some resistance $R$ (of the material it's made of - except for an ideal one..!) associated with it. Hence, Some work has to be done by external agencies in establishing current. This work done is stored as electromagnetic potential energy in an inductor. In case of capacitor, it's electrostatic..!

Induced emf, $e=-L\frac{dI}{dt}$ (negative sign indicates the opposing nature)

The small amount of work done over a small time $dt$ is $$dw=e.I.dt$$ $$\implies dw=-LI.dI$$ (As @Alfred said, Power $P=VI$ is used here...)

Thus, the total work done is establishing a steady current (say $I_o$) is $$W=-L\int_0^{I_0}I.dI=-\frac{1}2LI_o^2$$

Negative sign shows the opposing nature of the emf (a consequence of Lenz law) Thus, it has somewhat the same expression as that of a capacitor $\frac{1}2CV^2$. The small time $dt$ is taken into account because - Whenever you pass current through an inductor, steady current would be established only after some period of time. The current would be established in an increasing order. When the power supply is provided, the induced emf opposes the growth of current and when the power supply is cut-off, the emf now opposes the decay of current. (i.e. It goes in a decreasing order)

We could conclude that, the work done by these agencies is referred to as the energy stored in an inductor. Now, do you still relate both of them which are completely non-correlated..?

I think you're relating an Inductor coil with Magnetic Lorentz force.

Specifically, Lorentz force is experienced by a charged particle moving in an uniform magnetic field. $F=q(\vec{v}\times\vec{B})$. Let's see Inductor first...

Please keep in mind that an Inductor always has a self-inductance $L$ and some resistance $R$ (of the material it's made of - except for an ideal one..!) associated with it. Hence, Some work has to be done by external agencies in establishing current. This work done is stored as electromagnetic potential energy in an inductor.

Induced emf, $e=-L\frac{dI}{dt}$ (negative sign indicates the opposing nature, consequence of Lenz law)

The small amount of work done over a small time $dt$ is $$dw=e.I.dt$$ $$\implies dw=-LI.dI$$ (Power $P=VI$ is used here, because Power is the rate of doing work as we know...)

Thus, the total work done is establishing a steady current (say $I_o$) is $$W=-L\int_0^{I_0}I.dI=-\frac{1}2LI_o^2$$

Negative sign shows the opposing nature of the emf (same consequence of Lenz law). The small time $dt$ is taken into account because - Whenever you pass current through an inductor, steady current would be established only after some period of time. The current would be established in an increasing order. When the power supply is provided, the induced emf opposes the growth of current and when the power supply is cut-off, the emf now opposes the decay of current. (i.e. It goes in a decreasing order)

Hence, the work done by these agencies is referred to as the energy stored in an inductor.

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I think you're relating an Inductor coil with Magnetic Lorentz force.

Specifically, Lorentz force is experienced by a charged particle moving in an uniform magnetic field. $F=q(\vec{v}\times\vec{B})$. Let's see Inductor first...

Please keep in mind that an Inductor always have somehas a self-inductance $L$ associated with itand some resistance $R$ (exceptof the material it's made of - except for an ideal one..!) associated with it. Hence, Some work has to be done by external agencies in establishing current. This work done is stored as electromagnetic potential energy in an inductor. In case of capacitor, it's electrostatic..!

Induced emf, $e=-L\frac{dI}{dt}$ (negative sign indicates the opposing nature)

The small amount of work done over a small time $dt$ is $$dw=e.I.dt$$ $$\implies dw=-LI.dI$$ (As @Alfred said, Power $P=VI$ is used here...)

Thus, the total work done is establishing a steady current (say $I_o$) is $$W=-L\int_0^{I_0}I.dI=-\frac{1}2LI_o^2$$

Negative sign shows the opposing nature of the emf (a consequence of Lenz law) Thus, it has somewhat the same expression as that of a capacitor $\frac{1}2CV^2$. The small time $dt$ is taken into account because - Whenever you pass current through an inductor, steady current would be established only after some period of time. The current would be established in an increasing order. When the power supply is provided, the induced emf opposes the growth of current and when the power supply is cut-off, the emf now opposes the decay of current. (i.e. It goes in a decreasing order)

We could conclude that, the work done by these agencies is referred to as the energy stored in an inductor. Now, do you still relate both of them which are completely non-correlated..?

I think you're relating an Inductor coil with Magnetic Lorentz force.

Specifically, Lorentz force is experienced by a charged particle moving in an uniform magnetic field. $F=q(\vec{v}\times\vec{B})$. Let's see Inductor first...

Please keep in mind that an Inductor always have some self-inductance $L$ associated with it (except an ideal one..!). Hence, Some work has to be done by external agencies in establishing current. This work done is stored as electromagnetic potential energy in an inductor. In case of capacitor, it's electrostatic..!

Induced emf, $e=-L\frac{dI}{dt}$ (negative sign indicates the opposing nature)

The small amount of work done over a small time $dt$ is $$dw=e.I.dt$$ $$\implies dw=-LI.dI$$ (As @Alfred said, Power $P=VI$ is used here...)

Thus, the total work done is establishing a steady current (say $I_o$) is $$W=-L\int_0^{I_0}I.dI=-\frac{1}2LI_o^2$$

Negative sign shows the opposing nature of the emf (a consequence of Lenz law) Thus, it has somewhat the same expression as that of a capacitor $\frac{1}2CV^2$. The small time $dt$ is taken into account because - Whenever you pass current through an inductor, steady current would be established only after some period of time. The current would be established in an increasing order. When the power supply is provided, the induced emf opposes the growth of current and when the power supply is cut-off, the emf now opposes the decay of current. (i.e. It goes in a decreasing order)

We could conclude that, the work done by these agencies is referred to as the energy stored in an inductor. Now, do you still relate both of them which are completely non-correlated..?

I think you're relating an Inductor coil with Magnetic Lorentz force.

Specifically, Lorentz force is experienced by a charged particle moving in an uniform magnetic field. $F=q(\vec{v}\times\vec{B})$. Let's see Inductor first...

Please keep in mind that an Inductor always has a self-inductance $L$ and some resistance $R$ (of the material it's made of - except for an ideal one..!) associated with it. Hence, Some work has to be done by external agencies in establishing current. This work done is stored as electromagnetic potential energy in an inductor. In case of capacitor, it's electrostatic..!

Induced emf, $e=-L\frac{dI}{dt}$ (negative sign indicates the opposing nature)

The small amount of work done over a small time $dt$ is $$dw=e.I.dt$$ $$\implies dw=-LI.dI$$ (As @Alfred said, Power $P=VI$ is used here...)

Thus, the total work done is establishing a steady current (say $I_o$) is $$W=-L\int_0^{I_0}I.dI=-\frac{1}2LI_o^2$$

Negative sign shows the opposing nature of the emf (a consequence of Lenz law) Thus, it has somewhat the same expression as that of a capacitor $\frac{1}2CV^2$. The small time $dt$ is taken into account because - Whenever you pass current through an inductor, steady current would be established only after some period of time. The current would be established in an increasing order. When the power supply is provided, the induced emf opposes the growth of current and when the power supply is cut-off, the emf now opposes the decay of current. (i.e. It goes in a decreasing order)

We could conclude that, the work done by these agencies is referred to as the energy stored in an inductor. Now, do you still relate both of them which are completely non-correlated..?

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I think you're relating an Inductor coil with Magnetic Lorentz force.

Specifically, Lorentz force is experienced by a charged particle moving in an uniform magnetic field. $F=q(\vec{v}\times\vec{B})$. Let's see Inductor first...

Please keep in mind that an Inductor always have some self-inductance $L$ associated with it (except an ideal one..!). Hence, Some work has to be done by external agencies in establishing current. This work done is stored as electromagnetic potential energy in an inductor. In case of capacitor, it's electrostatic..!

Induced emf, $e=-L\frac{dI}{dt}$ (negative sign indicates the opposing nature)

The small amount of work done over a small time $dt$ is $$dw=e.I.dt$$ $$\implies dw=-LI.dI$$ (As @Alfred said, Power $P=VI$ is used here...)

Thus, the total work done is establishing a steady current (say $I_o$) is $$W=-L\int_0^{I_0}I.dI=-\frac{1}2LI_o^2$$

Negative sign shows the opposing nature of the emf (a consequence of Lenz law) Thus, it has somewhat the same expression as that of a capacitor $\frac{1}2CV^2$. The small time $dt$ is taken into account because - Whenever you pass current through an inductor, steady current would be established only after some period of time. The current would be established in an increasing order. When the power supply is provided, the induced emf opposes the growth of current and when the power supply is cut-off, the emf now opposes the decay of current. (i.e. It goes in a decreasing order)

We could conclude that, the work done by these agencies is referred to as the energy stored in an inductor. Now, do you still relate both of them which are completely non-correlated..?