Question 84: EM waves

Question

The oscillating magnetic field in a plane electromagnetic wave is given by $$B_{y}=5 	imes 10^{-6} \sin 1000 \pi\left(5 x-4 	imes 10^{8} tight) T$$. The amplitude of electric field will be :

Accepted Answer

In an electromagnetic wave, the magnitude of the electric field (E) and the magnetic field (B) are related by the speed of the wave (v), which can be represented as :
$$E = Bv$$
Here, B is the magnetic field, E is the electric field, and v is the speed of the electromagnetic wave.
In the given problem, the peak value of the magnetic field is given as :
$$B_0 = 5 	imes 10^{-6} T$$
The wave speed (v) can be calculated from the given equation for the magnetic field, using the relationship between frequency (f), wave number (k), and speed :
$$v = \frac{\omega}{k} = \frac{1000 \pi 	imes 4 	imes 10^8}{1000 \pi 	imes 5} = 8 	imes 10^7 m/s$$
Substituting these values into the equation for the electric field gives the peak value of the electric field :
$$E_0 = B_0 v = 5 	imes 10^{-6} T 	imes 8 	imes 10^7 m/s = 4 	imes 10^2 V/m$$

Answer

$$15 	imes 10^{2} \,\mathrm{Vm}^{-1}$$

Answer

$$5 	imes 10^{-6} \,\mathrm{Vm}^{-1}$$

Answer

$$16 	imes 10^{12} \,\mathrm{Vm}^{-1}$$

Answer

$$4 	imes 10^{2} \,\mathrm{Vm}^{-1}$$

EM waves

The oscillating magnetic field in a plane electromagnetic wave is given by $B_{y}=5 \times 10^{-6} \sin 1000 \pi\left(5 x-4 \times 10^{8} t\right) T$. The amplitude of electric field will be :

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