Question 153: Electromagnetic Induction

Question

Inductance of a coil with $10^4$ turns is 10 mH and it is connected to a dc source of 10 V with internal resistance of $10 \Omega$. The energy density in the inductor when the current reaches $\left(\frac{1}{e}ight)$ of its maximum value is $\alpha \pi 	imes \frac{1}{e^2} \mathrm{~J} / \mathrm{m}^3$. The value of $\alpha$ is
$\_\_\_\_$ .
$$ \left(\mu_0=4 \pi 	imes 10^{-7} \mathrm{Tm} / \mathrm{A}ight) . $$

Accepted Answer

When an inductor is connected to a DC voltage source (V) with a resistance (R), it initially opposes the flow of current. Over time, the current increases until it reaches a steady, maximum state.
Once the current is steady, the inductor acts like an ideal wire (zero resistance). Therefore, the maximum current ( $\mathrm{i}_{	ext {max }}$ ) is determined by Ohm's Law :
$$ \mathrm{i}_{\max }=\frac{\mathrm{V}}{\mathrm{R}} $$
For given circuit :
Voltage $\mathrm{V}=10 \mathrm{~V}$
Resistance $\mathrm{R}=10 \Omega$
$$ \mathrm{i}_{\max }=\frac{10}{10}=1 \mathrm{~A} $$
The current in the circuit is $\frac{1}{\mathrm{e}}$ time the maximum value, $\mathrm{i}=\mathrm{i}_{	ext {max }} 	imes\left(\frac{1}{\mathrm{e}}ight)$
$$ \mathrm{i}=1 	imes\left(\frac{1}{\mathrm{e}}ight)=\frac{1}{\mathrm{e}} \mathrm{~A} $$
For a long solenoid, the energy stored in an inductor is: $\mathrm{U}=\frac{1}{2} \mathrm{Li}^2$
The inductance of a long solenoid is: $\mathrm{L}=\frac{\mu_0 \mathrm{~N}^2 \mathrm{~A}}{\mathrm{l}}$, where N is total turns, A is cross-sectional area, and l is length.
The volume of the cylindrical core of the solenoid is: $\mathrm{V}_{\mathrm{vol}}=\mathrm{A} 	imes 1$
Then the energy density formula is $\mathrm{u}=\frac{\operatorname{energy}(\mathrm{U})}{\operatorname{volume}\left(\mathrm{V}_{	ext {vol }}ight)}$
$$ \mathrm{u}=\frac{\frac{1}{2} \mathrm{Li}^2}{\mathrm{~A} \cdot \mathrm{l}} $$
$$\Rightarrow $$ $u=\frac{\frac{1}{2}\left(\frac{\mu_0 N^2 A}{l}ight) i^2}{A \cdot l}$
$$\Rightarrow $$ $\mathrm{u}=\frac{\mu_0 \mathrm{~N}^2 \mathrm{i}^2}{2 \mathrm{l}^2}$
$$\Rightarrow $$ $$ \mathrm{u}=\frac{1}{2} \mu_0 \mathrm{n}^2 \mathrm{i}^2 $$
Where $\mathrm{n}=\frac{\mathrm{N}}{\mathrm{l}}$ is the number of turns per unit length.
Putting the given values into the derived energy density formula :
$$ \begin{aligned} \mu_0=4 \pi 	imes 10^{-7} \mathrm{Tm} / \mathrm{A}, \mathrm{n}=10^4 \mathrm{~m}^{-1}, \mathrm{i}=\frac{1}{\mathrm{e}} \mathrm{~A} ;\end{aligned} $$
$$\Rightarrow $$ $\mathrm{u}=\frac{1}{2} 	imes\left(4 \pi 	imes 10^{-7}ight) 	imes\left(10^4ight)^2 	imes\left(\frac{1}{\mathrm{e}}ight)^2$
$$\Rightarrow $$ $\mathrm{u}=20 \pi 	imes 1 / \mathrm{e}^2 \mathrm{~J} / \mathrm{m}^3$
$\Rightarrow 20 \pi 	imes \frac{1}{\mathrm{e}^2}=\alpha \pi 	imes \frac{1}{\mathrm{e}^2}$
$$ \Rightarrow \alpha=20 . $$
Therefore, the value of $\alpha$ is 20.

Electromagnetic Induction

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