At some instant and location, the electric field associated with an electromagnetic wave in vacuum has the strength 71.9 V/m. Find the magnetic field strength B, the total energy density u, and the power flow per unit area, all at the same instant and location.

a) Magnetic field strength, B = 2.397 * 10⁻⁷ T

b) Total energy density, U = 4.58 * 10⁻⁸ J/m³

c) Power flow per unit area, S = 13.71 W/m²

Explanation:

a) Electric field strength, E = 71.9 V/m

The relationship between the Electric field strength and the magnetic field strength in vacuum is:

E = Bc where c = 3.0 * 10⁸ m/s

71.9 = B * 3.0 * 10⁸

B = 71.9 / (3.0 * 10⁸)

B = 23.97 * 10⁻⁸

B = 2.397 * 10⁻⁷ T

b) Total Energy Density:

c)Power flow per unit area

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft