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babunello [35]
4 years ago
10

Astar begins as a blank a large cloud of gas and dust

Physics
1 answer:
crimeas [40]4 years ago
3 0
This is a statement but yes a star forms inside nebulae which are gigantic clouds of gas. stars form inside as the gases own gravity pulls it together after which it becomes large enough to perform fusion and become a star.
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Motion

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Which of the following expressions gives the ratio of the energy density of the magnetic field to that of the electric field jus
miss Akunina [59]

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;

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Energy density of electric field

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

where;

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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

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Nookie1986 [14]

Answer:

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