Question

A certain 60.0 Hz AC power line radiates an electromagnetic wave having a maximum electric field strength of 11.6 kV/m.

Answer 1

Explanation:

Given that,

Frequency of the power line, f = 6 Hz

Value of maximum electric field strength of 11.6 kV/m

(a) The wavelength of this very low frequency electromagnetic wave is given by using relation as :

$c=f\lambda$

$\lambda=\dfrac{c}{f}$

$\lambda=\dfrac{3\times 10^8\ m/s}{60\ Hz}$

$\lambda=5\times 10^6\ m$

(b) As its can be seen that the wavelength of this wave is very high. It shows that it is a radio wave.

(c) The relation between the maximum magnetic field strength and maximum electric field strength is given by :

$B_0=\dfrac{E_0}{c}\\\\B_0=\dfrac{11.6\times 10^3}{3\times 10^8}\\\\B_0=3.86\times 10^{-5}\ T$

So, the maximum magnetic field strength is $3.86\times 10^{-5}\ T$.

Answer 2

Answer:

a) $\lambda=5\times 10^6\ m$

b) The wavelength of the obtained is greater than 1mm so it lies in the range of radio waves.

c) $B_m=3.867\times 10^{-5}\ T$

Explanation:

Given:

frequency of electromagnetic waves, $f=60\ Hz$

maximum field strength of the electric field, $E_m=11600\ V.m^{-1}$

Since the velocity of electromagnetic waves is, $c=3\times 10^8\ m.s^{-1}$

a)

We know the relation between the frequency and wavelength is given as:

$\lambda=\frac{c}{f}$

$\lambda=\frac{3\times 10^{8}}{60}$

$\lambda=5\times 10^6\ m$

b)

The wavelength of the obtained is greater than 1mm so it lies in the range of radio waves.

c)

The maximum magnetic field can be calculated as:

$B_m=\frac{E_m}{c}$

$B_m=\frac{11600}{3\times 10^8}$

$B_m=3.867\times 10^{-5}\ T$