Question

Electric field of electromagnetic wave is $E = 800 \sin \pi \left( 10^8 t + \frac{x}{150} \right) \text{ V/m}$. Where x in cm and 't' in sec. A charged particle is moving with speed of $1.5 \times 10^6 \text{ m/s}$. Find the ratio of magnetic force to electric force on charge particle.

• A. $10^{+2}$
• B. $10^{-2}$
• C. $2 \times 10^2$
• D. $2 \times 10^{-2}$

Answer 1

The Correct Option is B

Step 1: Calculate the wave velocity $v_w$ from the equation.
$v_w = \frac{\omega}{k} = \frac{10^8 \pi}{\pi/150} = 1.5 \times 10^8 \text{ m/s}$ (after conversion from cm).

Step 2: Express the ratio of forces.
Electric force $F_E = qE$.
Magnetic force $F_M = qvB$.
In an EM wave, $E = v_w B$.
Ratio $\frac{F_M}{F_E} = \frac{qvB}{qv_w B} = \frac{v}{v_w}$.

Step 3: Substitute given particle velocity $v = 1.5 \times 10^6 \text{ m/s}$.
Ratio = $\frac{1.5 \times 10^6}{1.5 \times 10^8} = 10^{-2}$.

Final Answer: Option (2).