Energy Stored in an Inductor: Formula, Derivation & Example Problems

22.4 ENERGY STORED IN AN INDUCTOR

22.4.1 Energy Stored in an Inductor

Figure 22.62 The figure shows a series circuit consisting of an ohmic resistor (R), an ideal inductor (L), and a battery of emf (E) with negligible internal resistance. When the switch is closed, current begins to flow and increases with time due to the effect of self-inductance in the inductor.

1. An ohmic resistor of resistance R, an ideal inductor of self inductance L and a battery of emf E and negligible resistance are connected in series as shown in Figure 22.62. When the switch is closed, current starts to flow in the circuit. Let the instantaneous current at time t be i.
2. Applying Kirchhoff’s voltage law to the circuit, we get
   $E = iR + L \left(\frac{di}{dt}\right)$
   where di/dt is the rate of change of current at time t.
3. The power P supplied by the battery to the resistor and inductor at time t is
   $P = P_R + P_L$
   $iE = i(iR) + iL \left(\frac{di}{dt}\right)$

4. The electrical power $P_L$ received by the inductor at time t is

   $P_L=iL \frac{di}{dt}$
5. The total energy U_L received by the inductor when the current has just reached a maximum constant value I at time T is
   $U_L=\int_{0}^{T}P_L dt$

   
   

   $=\int_{0}^{I}\left(Li\frac{di}{dt}\right)$

   
   

   $=L\left[\frac{1}{2}i^2\right]_{0}^{I}$

   
   

   $U_L=\frac{1}{2}LI^2$
              This amount of energy is stored by the inductor in the magnetic field which is produced by the current I.
6. When the circuit is switched off, the current starts to drop to zero. Simultaneously, the magnetic field shrinks and finally disappears when the current becomes zero. The energy previously stored in the magnetic field is converted back to electrical energy, which reappears in the circuit.

EXAMPLE 22.32

Explain the terms self induction and self inductance of a circuit or circuit component. Derive an expression for the self inductance of a long solenoid. Give one main reason why this expression only gives an approximate value for the self inductance of the solenoid. 

A solenoid of length 550 cm and diameter 5.0 cm has 1 000 turns. A constant current of 2.5 A flows through the solenoid, which can be assumed to behave as an ideal inductor. Determine

(a) the self inductance of the solenoid

(b) the energy stored in the magnetic field produced by the current

(c) the average back emf produced in the solenoid if the circuit is switched off and the current drops to zero in 10 ms.

Answer

(a) The self inductance L of the solenoid is given by

$L=\mu_0\left(\frac{N^2}{l}\right)A$

$=(4\pi \times 10^{-7})\left(\frac{1000^2}{0.50}\right)\left(\frac{1}{4}\pi \times 0.050^2\right)$

$=4.93 \times 10^{-3} \ H$

(b) The energy stored by solenoid is

$U_L=\frac{1}{2}LI^2$

$=\frac{1}{2}(4.93 \times 10^{-3})(2.5)^2=15.4 \ mJ$

(c) The average induced emf

$E_b= −L \left(\frac{\Delta I}{\Delta t}\right)$

$=-(-4.93 \times 10^{-3})\left(\frac{0-2.5}{10\times 10^{-3}}\right)=1.2 \ V$

EXAMPLE 22.33

Figure 22.63 The figure shows a circuit containing an ideal inductor in which a steady current flows. When the current is constant, energy is stored in the magnetic field of the inductor. When the circuit is switched off, the current decreases to zero, producing a back emf across the inductor due to the change in current.

Refer to the circuit shown in Figure 22.63 Determine
(a) the energy stored by the ideal inductor when the current is constant
(b) the average back emf produced across the inductor if the circuit is switched off and the constant current drops to zero in 10 ms.

Answer

(a) Since the inductor is ideal, it does not have resistance. The constant current I is given by

$I = \frac{E}{r + R}$

$= \frac{12}{2 + 10} = 1.0 \ A$

The energy stored by the inductor is

$U_L = \frac{1}{2} LI^2$

$= \frac{1}{2} (0.250)(1.0)^2= 0.13 \ J$

(b) The average induced emf

$E_b= −L \left(\frac{\Delta I}{\Delta t}\right)$

$=-(-0.250)\left(\frac{0-1}{10\times 10^{-3}}\right)=25 \ V$

EXAMPLE 22.34

At the instant when a current of 3.5 A flows through an ideal 200 mH inductor, the current is increasing at a rate of 10 A s⁻¹. At that particular instant, determine
(a) the magnetic flux linkage through the inductor
(b) the back emf produced across the inductor
(c) the rate at which energy is supplied to the inductor

Answer

(a)
$LI = N\Phi$ 

= flux linkage

Hence, flux linkage $= LI$

$= (0.200)(3.5) = 0.70 \ Wb$

(b) Back emf 

$E_b= −L \left(\frac{\Delta I}{\Delta t}\right)$

$= −(0.200)(+10) = −2.0 \ V$

(c) The instantaneous power supply is

$P_L=LI \frac{dI}{dt}$

$= (0.200)(3.5)(10) = 7.0 \ W$

Applications of Energy Stored in an Inductor

⚡ Power Supplies

Inductors store and release energy to maintain stable voltage in DC-DC converters.

🔊 Audio Systems

Used in speakers and filters to control frequencies and improve sound quality.

📡 Radio Circuits

Inductors help tuning circuits select specific frequencies in communication systems.

🚗 Electric Vehicles

Used in motor controllers to regulate current and improve efficiency.

⚙️ Transformers

Transfer energy between coils using magnetic fields in power systems.

💡 Energy Storage

Stores energy temporarily in magnetic fields for switching circuits.

Conclusion

The energy stored in an inductor is an important concept in electromagnetism, where energy is stored in the magnetic field created by electric current. This energy depends on the inductance and the square of the current flowing through the circuit.

Inductors play a vital role in many real-world applications such as power supplies, transformers, communication systems, and electric vehicles. By understanding how energy is stored and released, we can better analyze and design electrical and electronic systems.

Overall, mastering this concept helps students build a strong foundation in physics and prepares them for solving practical and advanced problems in electricity and magnetism.

Frequently Asked Questions (FAQ)

What is the energy stored in an inductor?

The energy stored in an inductor is the energy kept in its magnetic field when current flows through it.

What is the formula for energy stored in an inductor?

The formula is U = ½LI², where L is inductance and I is the current.

Why does an inductor store energy?

An inductor stores energy because a changing current creates a magnetic field, which stores energy around the coil.

Where is energy stored in an inductor?

The energy is stored in the magnetic field surrounding the inductor.

What are the applications of inductors?

Inductors are used in power supplies, transformers, radio circuits, and energy storage systems.

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