Question

What is the magnetic potential energy stored in a cylindrical volume of height hcylin = 50 mm and radius Rcylin = 24 mm that symmetrically surrounds an infinitely long wire that has radius Rwire = 2.1 mm and carries current I = 2.9 A ? The volume in which the energy should be calculated does not contain the wire

Answer 1

Answer:

$U=1.02\times 10^{-7}\ J$

Explanation:

Given that

h= 50 mm = 0.05 m

R= 24 mm = 0.024 m

Rw= 2.1 mm = 0.0021 m

I= 2.9 A

We know that magnetic filed given as

$B=\dfrac{\mu _oI}{2\pi r}$

Volume of small element

dV= 2πr hdr

Now the magnetic potential energy given as

$U=\int \dfrac{B^2}{2\mu_o}dV$

$B=\dfrac{\mu _oI}{2\pi r}$

dV= 2πr hdr

$U=\int\dfrac{\left(\dfrac{\mu _oI}{2\pi r}\right)^2}{2\mu_o}2\pi rhdr$

$U=\dfrac{\mu _oI^2h}{4\pi}\int_{R_w}^{R}\dfrac{dr}{r}$

$U=\dfrac{\mu _oI^2h}{4\pi}\times \ln\dfrac{R}{R_w}$

Now by putting the values

$U=\dfrac{4\pi \times 10^{-7}\times 2.9^2\times 0.05}{4\pi}\times \ln\dfrac{0.024}{0.0021}\ J$

$U=1.02\times 10^{-7}\ J$