A carriage of 20 kg is pulled with a force of 35 N. How far the carriage will go
1 answer:
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Answer:
electric field amplitude is 0.1133 V/m
Explanation:
given data
energy density = 5.69 × 10^−14 J/m3
speed of light = 2.99792 × 10^8 m/s
permeability of free space = 4π × 10^−7 N/A2
to find out
corresponding electric field amplitude
solution
we know electric filed amplitude E is
E = BC ..............1
so first we find magnetic filed B from energy density
that is energy density
u = B²/ 2µ
so B = √2µu
put value
B = √2(4π××5.69 ×
)
B = 3.780645 ×
so from equation 1
E = 3.780645 × (2.99792 × 10^8)
E = 0.1133
electric field amplitude is 0.1133 V/m