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

To determine the ratio of the magnetic force to the electric force on a charged particle, let's break down the given problem step-by-step.

The electric field of the electromagnetic wave is given by:

\(E = 800 \sin \pi \left( 10^8 t + \frac{x}{150} \right) \text{ V/m}\)

A charged particle is moving with a velocity \(v = 1.5 \times 10^6 \text{ m/s}\). We need to find the ratio of the magnetic force to the electric force on this particle.

Step 1: Determine the Magnetic Field

The relation between the electric field \(E\) and the magnetic field \(B\) in an electromagnetic wave is given by:

\(c = \frac{E}{B}\)

where \(c\) is the speed of light, approximately \(3 \times 10^8 \text{ m/s}\).

Rearranging the equation, we have:

\(B = \frac{E}{c}\)

Given \(E = 800 \text{ V/m}\), the magnetic field \(B\) is:

\(B = \frac{800}{3 \times 10^8} \text{ T}\)

\(B = \frac{8}{3} \times 10^{-6} \text{ T}\)

Step 2: Calculate Magnetic Force and Electric Force

The magnetic force \(F_B\) on a charge \(q\) moving with velocity \(v\) in a magnetic field \(B\) is given by:

\(F_B = qvB \sin \theta\)

Assuming the maximum force scenario (i.e., \(\theta = 90^\circ\), so \(\sin \theta = 1\)), the magnetic force is:

\(F_B = qvB\)

The electric force \(F_E\) on a charge \(q\) in an electric field \(E\) is:

\(F_E = qE\)

Step 3: Calculate the Ratio

The ratio of the magnetic force to the electric force is:

\(\frac{F_B}{F_E} = \frac{qvB}{qE} = \frac{vB}{E}\)

Substitute the values:

\(\frac{F_B}{F_E} = \frac{1.5 \times 10^6 \times \frac{8}{3} \times 10^{-6}}{800}\)

Calculate the above expression:

\(\frac{F_B}{F_E} = \frac{1.5 \times 8}{3 \times 800}\) \(\frac{F_B}{F_E} = \frac{12}{2400} = \frac{1}{200} = 10^{-2}\)

Therefore, the ratio of magnetic force to electric force on the charged particle is \(10^{-2}\).