Answer:
[tex](d) \ \ \frac{\mu_o}{\epsilon_o} (\frac{L}{2\pi r*R} )^2[/tex]
Explanation:
Energy density in magnetic field is given as;
[tex]U_B = \frac{1}{2 \mu_o} B^2[/tex]
where;
B is the magnetic field strength
Energy density of electric field
[tex]U_E = \frac{1}{2}\epsilon E^2[/tex]
where;
E is electric field strength
Take the ratio of the two fields energy density
[tex]\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})[/tex]
[tex]\frac{U_B}{U_E} = \frac{1}{\mu_o \epsilon} (\frac{B}{E})^2[/tex]
But, Electric field potential, V = E x L = IR (I is current and R is resistance)
[tex]\frac{U_B}{U_E} = \frac{1}{\mu_o \epsilon} (\frac{B*L}{E*L})^2[/tex]
Now replace E x L with IR
[tex]\frac{U_B}{U_E} = \frac{1}{\mu_o \epsilon} (\frac{B*L}{IR})^2[/tex]
Also, B = μ₀I / 2πr, substitute this value in the above equation
[tex]\frac{U_B}{U_E} = \frac{1}{\mu_o \epsilon} (\frac{\mu_oI*L}{2\pi r* IR})^2[/tex]
cancel out the current "I" and factor out μ₀
[tex]\frac{U_B}{U_E} = \frac{\mu_o^2}{\mu_o \epsilon} (\frac{L}{2\pi r* R})^2[/tex]
Finally, the equation becomes;
[tex]\frac{U_B}{U_E} = \frac{\mu_o}{\epsilon} (\frac{L}{2\pi r*R })^2[/tex]
Therefore, the correct option is (d) μ₀/ϵ₀ (L /R 2πr)²
