Question

Which of the following expressions gives the ratio of the energy density of the magnetic field to that of the electric field just inside the surface of a wire of length L, radius r, and resistance R which is carrying a current I?
a) μ₀/ϵ₀ (I L /R 2πr)²
b) 1/μ₀ ϵ₀ (I L /R 2πr)²
c) 1/μ₀ ϵ₀ (L /R 2πr)²
d) μ₀/ϵ₀ (L /R 2πr)²

Answer 1

Answer:

$(d) \ \ \frac{\mu_o}{\epsilon_o} (\frac{L}{2\pi r*R} )^2$

Explanation:

Energy density in magnetic field is given as;

$U_B = \frac{1}{2 \mu_o} B^2$

where;

B is the magnetic field strength

Energy density of electric field

$U_E = \frac{1}{2}\epsilon E^2$

where;

E is electric field strength

Take the ratio of the two fields energy density

$\frac{U_B}{U_E} = \frac{1}{2\mu_o} B^2 / \frac{1}{2}\epsilon E^2\\\\\frac{U_B}{U_E} = \frac{B^2}{2\mu_o} *\frac{2}{\epsilon E^2} \\\\\frac{U_B}{U_E} = \frac{1}{\mu_o \epsilon} (\frac{B^2}{E^2})$

$\frac{U_B}{U_E} = \frac{1}{\mu_o \epsilon} (\frac{B}{E})^2$

But, Electric field potential, V = E x L = IR (I is current and R is resistance)

$\frac{U_B}{U_E} = \frac{1}{\mu_o \epsilon} (\frac{B*L}{E*L})^2$

Now replace E x L with IR

$\frac{U_B}{U_E} = \frac{1}{\mu_o \epsilon} (\frac{B*L}{IR})^2$

Also, B = μ₀I / 2πr, substitute this value in the above equation

$\frac{U_B}{U_E} = \frac{1}{\mu_o \epsilon} (\frac{\mu_oI*L}{2\pi r* IR})^2$

cancel out the current "I" and factor out μ₀

$\frac{U_B}{U_E} = \frac{\mu_o^2}{\mu_o \epsilon} (\frac{L}{2\pi r* R})^2$

Finally, the equation becomes;

$\frac{U_B}{U_E} = \frac{\mu_o}{\epsilon} (\frac{L}{2\pi r*R })^2$

Therefore, the correct option is (d) μ₀/ϵ₀ (L /R 2πr)²