An 800 kHz radio signal is detected at a point 3.2 km distant from a transmitter tower. The electric field amplitude of the signal at that point is 320 mV/m. Assume that the signal power is radiated uniformly in all directions and that radio waves incident upon the ground are completely absorbed. The magnetic field amplitude of the signal at that point, in nT, is closest to:

1.07 nT

Explanation:

We know that E/B = c where E = electric field amplitude = 320 mV/m = 0.32 V/m, B = magnetic field amplitude and c = speed of light = 3 × 10⁸ m/s.

So, B = E/c

Substituting E and c into B, we have

B = E/c

= 0.32 V/m ÷ 3 × 10⁸ m/s

= 0.1067 × 10⁻⁸ T

= 1.067 × 10⁻⁹ T

= 1.067 nT

≅ 1.07 nT

not too sure thinking it will always be directed towards the centre of the earth. could be wrong not sure

electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j