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3 years ago

5

On a clear day at a certain location, a 119-V/m vertical electric field exists near the Earth's surface. At the same place, the

Earth's magnetic field has a magnitude of 5.050 10-5 T.
(a) Compute the energy density of the electric field. nJ/m3
(b) Compute the energy density of the magnetic field. µJ/m3

1 answer:

IrinaVladis [17]3 years ago

8 0

Answer:

(a) 62.69 nJ/m^3

(b) 1015.22 μJ/m^3

Explanation:

Electric field, E = 119 V/m

Magnetic field, B = 5.050 x 10^-5 T

(a) Energy density of electric field = $\frac{1}{2}\varepsilon _{0}E^{2}$

$=\frac{1}{2}\times 8.854\times 10^{-12}\times 119\times 119$

= 6.269 x 10^-8 J/m^3 = 62.69 nJ/m^3

(b) energy density of magnetic field = $\frac{B^{2}}{2\mu _{0}}$

$=\frac{\left ( 5.05\times 10^{-5} \right )^{2}}{2\times 4\times 3.14\times 10^{-7}}$

= 1.01522 x 10^-3 J/m^3 = 1015.22 μJ/m^3

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