Asinusoidal electromagnetic wave is propagating in a vacuum in the +z-direction.

part a

if at a particular instant and at a certain point in space the electric field is in the +x-direction and has a magnitude of 3.40v/m , what is the magnitude of the magnetic field of the wave at this same point in space and instant in time?

part b

what is the direction of the magnetic field?

a)1.13 10-8 T.  b) +y direction

Explanation:

a)

For an electromagnetic wave propagating in a vacuum, the wave speed is c = 3. 108 m/s.

At a long distance from the source, the components of the wave (electric and magnetic fields) can be considered as plane waves, so the equations for them can be written as follows:

E(z,t) = Emax cos (kz-ωt-φ) +x

B(z,t) = Bmax cos (kz-ωt-φ) +y

In an electromagnetic wave, the magnetic field and the electric field, at any time, and at any point in space, as the perturbation is propagating at a speed equal to c (light speed in vacuum), are related by this expression:

Bmax = Emax/c

So, solving for Bmax:

Bmax = 3.4 V/m / 3 108 m/s = 1.13 10-8 T.

b) As we have already said, in an electromagnetic wave, the electric field and the magnetic field are perpendicular each other and to the propagation direction, so in this case, the magnetic field propagates in the +y direction.

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

i believe the answer is electric potential.

explanation: