Problem 2143 - JEE Physics

Question

A plane electromagnetic wave of frequency \(25 \mathrm{MHz}\) travels in free space along the \(\mathrm{x}\) direction. At a particular point in space and time \(\mathrm{E}^{-}=6.3 \mathrm{j} \wedge \mathrm{Vm}^{-1}\) then \(\mathrm{B}^{-}\) at this point is (A) \(2.1 \times 10^{-8}\) i \(\mathrm{T}\) (B) \(2.1 \times 10^{-8} \mathrm{k} \wedge \mathrm{T}\) (C) \(1.89 \times 10^{9} \mathrm{k} \wedge \mathrm{T}\) (D) \(2.52 \times 10^{-7} \mathrm{k} \wedge \mathrm{T}\)

Step-by-Step Solution

Answer

The short answer based on the provided step-by-step solution is: The magnetic field strength at this point is \(B = 2.1 \times 10^{-8} k \frac{T}{m}\).

Step 1: 1. Write down the given information

The given information is: frequency \(f = 25 MHz\), electric field strength \(E = 6.3j \frac{V}{m}\), and the wave is traveling in the x-direction.

Step 2: 2. Convert frequency to angular frequency

We will convert the frequency, \(f\), to angular frequency, \(\omega\), using the formula: \[\omega = 2 \pi f\] Plugging in the given frequency, we get: \[\omega = 2 \pi (25 \times 10^6 Hz)\]

Step 3: 3. Relationship between electric and magnetic fields

The relationship between the electric field, E, and the magnetic field, B, in an electromagnetic wave is given by: \[E = cB\] where \(c\) is the speed of light in free space.

Step 4: 4. Calculate the magnetic field strength

Now, we can solve for the magnetic field strength, B, using the given electric field strength, E, and the speed of light, c: \[B = \frac{E}{c}\] Plugging in the values, we get: \[B = \frac{6.3j \frac{V}{m}}{3 \times 10^8 \frac{m}{s}}\]

Step 5: 5. Simplify the expression to get the final answer

Simplifying the expression above, we get: \[B = 2.1 \times 10^{-8} j \frac{T}{m}\] Since the wave is traveling in the x-direction, our result should be in the k direction (z-direction), so the answer is: \[B = 2.1 \times 10^{-8} k \frac{T}{m}\] Comparing our result with the given options, the correct answer is (B) \(2.1 \times 10^{-8} k \frac{T}{m}\).

Key Concepts

Angular Frequency

Angular frequency is a key concept when discussing waves, especially electromagnetic waves. It is often represented by the Greek letter \( \omega \). Angular frequency essentially describes how fast the wave oscillates in radians per second. To calculate angular frequency, we use the formula:

• \(\omega = 2\pi f\)

where \(f\) is the frequency in hertz (Hz). For an electromagnetic wave with a frequency of 25 MHz, the angular frequency becomes:

• \(\omega = 2\pi (25 \times 10^6 Hz)\)

This conversion from frequency to angular frequency is crucial in connecting how often a wave cycle completes with how swiftly it travels through space.Understanding angular frequency helps us analyze the dynamic properties of waves, such as their speed and the relationship between electric and magnetic fields.

Electric Field

The electric field is a vector field that represents the magnitude and direction of the force that would be exerted on a positive test charge placed in the field. In the context of electromagnetic waves, like the one in this exercise, the electric field component is time-varying and related to the wave's propagation through space.For instance, in this problem, the electric field at a specific point in space is given as:

• \(E = 6.3\, \hat{j} \frac{V}{m}\)

This indicates the strength of the electric field is constant at 6.3 V/m and is oriented in the y-direction (indicated by \(\hat{j}\)).The electric field is crucial for determining the magnetic field in electromagnetic waves, as indicated by the formula \(E = cB\), where \(c\) is the speed of light.

Magnetic Field

The magnetic field component of an electromagnetic wave is crucial for understanding how energy is transported through space. It is measured in teslas (T).In an electromagnetic wave, the electric field and the magnetic field are perpendicular to each other and also to the direction of the wave's travel.In this exercise, after calculating, the magnetic field \(B\) at a specific point is found to be along the z-direction (or \(\hat{k}\) direction), calculated as:

• \(B = 2.1 \times 10^{-8} \hat{k} \frac{T}{m}\)

This was determined using the relationship between the electric field \(E\) and the speed of light \(c\):

• \(B = \frac{E}{c}\)

This relationship tells us that the magnetic field is directly proportional to the electric field for electromagnetic waves, governed by the speed of light.

Speed of Light

The speed of light in free space is a fundamental constant of nature, denoted \(c\), with a value of approximately \(3 \times 10^8 \frac{m}{s}\). This speed sets a universal limit and plays a vital role in the propagation of electromagnetic waves. For electromagnetic waves, the speed of light connects the electric field \(E\) and the magnetic field \(B\) through the equation:

• \(E = cB\)

This indicates that the energy and information carried by electromagnetic waves rely on this constant speed.In the context of this problem, using the speed of light allowed us to calculate the magnetic field from the given electric field:

• \(B = \frac{E}{c}\)

Using the speed of light in these calculations helps in accurately relating the wave's electric and magnetic components and their propagation through space, highlighting the essential nature of this speed in electromagnetic theory.