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Electromagnetic Induction — Physics STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 SciencePhysicsElectromagnetic Induction3 Marks
Question
A coaxial cable, whose outer radius is five times its inner radius, is carrying a current of $1.5 A$. What is the magnetic field energy stored in a $2 m$ length of the cable?
✓

Answer

Data $: b / a=5, I=1.5 A , $
$l =2 m , \frac{\mu_0}{4 \pi}=10^{-7} H / m$
The total magnetic energy in a given length of a current $-$ carrying coaxial cable,
$U _{ m }=\left(\frac{\mu_0}{4 \pi}\right) I^2 l \log _e \frac{b}{a}$
Therefore, the required magnetic energy is
$U_m=\left(10^{-7}\right)(1.5)^2(2) \log _e 5$
$=4.5 \times 10^7 \times 2.303 \times \log _{10} 5$
$=4.5 \times 10^{-7} \times 2.303 \times 0.6990$
$=7.24 \times 10^{-7} J$

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