\(E_y=60\sin[\frac{π}{4}×10^3(x−3×10^8t)]\hat{j}Vm^{−1} \)
\(B_y = 2\sin\left[\frac{\pi}{4} \times 10^3 (x - 3 \times 10^8 t)\right]\hat{k}T\)
\(E_y = 60\sin\left[\frac{\pi}{4} \times 10^3 (x - 3 \times 10^8 t)\right] \hat{j} \,Vm^{-1}\)
\(B_z = 2 \times 10^{-7} \sin\left[\frac{\pi}{4} \times 10^3 (x - 3 \times 10^8 t)\right] \hat{k} \, \text{T}\)
\(E_y = 2 \times 10^{-7} \sin\left(\frac{\pi}{4} \times 10^3 (x - 3 \times 10^8 t)\right) \hat{j} \, Vm^{-1}\)
\(B_z = 60 \sin\left[\frac{\pi}{4} \times 10^3 (x - 3 \times 10^8 t)\right]\hat{k} \, \text{T}\)
\(E_y = 2 \times 10^{-7} \sin\left[\frac{\pi}{4} \times 10^4 (x - 4 \times 10^8 t)\right] \hat{j} \, Vm^{-1}\)
\(B_z = 60 \sin\left[\frac{\pi}{4} \times 10^4 x - 4 \times 10^8 t)\right] \hat{k} \, \text{T}\)
The correct answer is (B) : \(E_y = 60\sin\left[\frac{\pi}{4} \times 10^3 (x - 3 \times 10^8 t)\right] \hat{j} \,Vm^{-1}\)
\(B_z = 2 \times 10^{-7} \sin\left[\frac{\pi}{4} \times 10^3 (x - 3 \times 10^8 t)\right] \hat{k} \, \text{T}\)
In first 3 options speed of light is 3 × 108 m/sec and in the fourth option it is 4 × 108 m/sec.
Using
E = CB
We can check the option is B.
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is:
The waves that are produced when an electric field comes into contact with a magnetic field are known as Electromagnetic Waves or EM waves. The constitution of an oscillating magnetic field and electric fields gives rise to electromagnetic waves.
Electromagnetic waves can be grouped according to the direction of disturbance in them and according to the range of their frequency. Recall that a wave transfers energy from one point to another point in space. That means there are two things going on: the disturbance that defines a wave, and the propagation of wave. In this context the waves are grouped into the following two categories: