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(A) The energy stored in a solenoid is given by $U = \frac{1}{2} LI^2$.The self-inductance $L$ of a solenoid is $L = \frac{\mu_0 \mu_r N^2 A}{\ell}$,

Vedclass · Accepted Answer

$15.7$

Vedclass · Answer

(A) The energy stored in a solenoid is given by $U = \frac{1}{2} LI^2$.The self-inductance $L$ of a solenoid is $L = \frac{\mu_0 \mu_r N^2 A}{\ell}$, where $N$ is the total number of turns and $\ell$ is the length.Energy per unit length $u = \frac{U}{\ell} = \frac{1}{\ell} \left( \frac{1}{2} \frac{\mu_0 \mu_r N^2 A}{\ell} I^2 ight) = \frac{1}{2} \mu_0 \mu_r \left( \frac{N}{\ell} ight)^2 A I^2$.Given: turn density $n = \frac{N}{\ell} = 5000 \, 	ext{turns/m}$, area $A = 10 \, 	ext{cm}^2 = 10 	imes 10^{-4} \, 	ext{m}^2$, current $I = 1 \, 	ext{A}$, relative permeability $\mu_r = 1000$, and $\mu_0 = 4\pi 	imes 10^{-7} \, 	ext{T} \cdot 	ext{m/A}$.Substituting the values:$u = \frac{1}{2} 	imes (4\pi 	imes 10^{-7}) 	imes 1000 	imes (5000)^2 	imes (10 	imes 10^{-4}) 	imes (1)^2$$u = \frac{1}{2} 	imes 4\pi 	imes 10^{-7} 	imes 10^3 	imes 25 	imes 10^6 	imes 10^{-3} 	imes 1$$u = 2\pi 	imes 10^{-7} 	imes 25 	imes 10^9 	imes 10^{-3} = 50\pi \approx 157.08 	imes 0.1 = 15.7 \, 	ext{J/m}$.

$A$ solenoid has a turn density of $5000 \, \text{turns/m}$ and a cross-sectional area of $10 \, \text{cm}^2$. If a current of $1 \, \text{A}$ flows through it and the core material has a relative permeability of $1000$, find the energy per unit length of the solenoid in $\text{J/m}$.

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