Problem 2144 - 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) \(2.1 \times 10^{-8} k \frac{T}{m}\).