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
Answer:
The magnetic field will help to create an electromagnetic radiation which will prevent the charged particle from moving in a constant direction in a straight path.
Explanation: A magnetic field is an area around a particle where magnetic energy is felt or experienced, this ensures that all the charged particle within the magnetic field will be influenced by magnet.
In the highlighted situation, the magnetic field will help to ensure that an electromagnetic radiation is created to synchrotron radiation occurs effectively.
Using kinematics: 8.1 = 1/2(9.8)(t^2),
t = 1.2857 s.
Horizontal distance x travelled is x = vhorizontal * t, so 9.3 = v*1.2857, or v= 7.233 m/s horizontally.