The electric field a plane electromagnetic wave is given by : E_y = 69 sin [ 0.6 times 10^3 x - 1.8 times 10^11 t ]text V/m.$$E_y = 69 \sin [ 0.6 \times 10^3 x - 1.8 \times 10^{11} t ]\text{ V/m}.$$ The expression for magnetic field associated with this electromagnetic wave is ____ T.

### Related Formula B_0 = fracE_0c$$B_0 = \frac{E_0}{c}$$ hatc = hatE times hatB$$\hat{c} = \hat{E} \times \hat{B}$$ ### Core Logic The phase of the wave is (0.6 times 10^3 x - 1.8 times 10^11 t)$(0.6 \times 10^3 x - 1.8 \times 10^{11} t)$. This indicates the wave propagates in the +x$+x$ direction, so hatc = hati$\hat{c} = \hat{i}$. The electric field oscillates along the y$y$-axis, so hatE = hatj$\hat{E} = \hat{j}$. From hatB = hatc times hatE$\hat{B} = \hat{c} \times \hat{E}$, we have hatB = hati times hatj = hatk$\hat{B} = \hat{i} \times \hat{j} = \hat{k}$. So, the magnetic field is along the z$z$-axis (B_z$B_z$). ### Step 1: Calculate Amplitude of B Wave speed v = c = fracomegak = frac1.8 times 10^110.6 times 10^3 = 3 times 10^8text m/s$v = c = \frac{\omega}{k} = \frac{1.8 \times 10^{11}}{0.6 \times 10^3} = 3 \times 10^8\text{ m/s}$. The amplitude of the magnetic field is: B_0 = fracE_0c = frac693 times 10^8 = 23 times 10^-8 = 2.3 times 10^-7text T$$B_0 = \frac{E_0}{c} = \frac{69}{3 \times 10^8} = 23 \times 10^{-8} = 2.3 \times 10^{-7}\text{ T}$$ The phase remains exactly the same as the electric field: B_z = 2.3 times 10^-7 sin(0.6 times 10^3 x - 1.8 times 10^11 t)$$B_z = 2.3 \times 10^{-7} \sin(0.6 \times 10^3 x - 1.8 \times 10^{11} t)$$ ### Pattern Recognition B_0 = E_0/c$B_0 = E_0/c$ gives the magnitude. The vector identity hatB = hatv times hatE$\hat{B} = \hat{v} \times \hat{E}$ gives the direction. Phase part never changes sign or terms between E and B equations. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electromagnetic Waves