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Physics Question on mechanical properties of solids

A plane electromagnetic wave of frequency 35 MHz travels in free space along the X-direction. At a particular point (in space and time), E⃗=9.6j^ V/m\vec{E} = 9.6 \hat{j} \, \text{V/m}. The value of the magnetic field at this point is:

A

3.2×10−8k^ T3.2 \times 10^{-8} \hat{k} \, \text{T}

B

3.2×10−8i^ T3.2 \times 10^{-8} \hat{i} \, \text{T}

C

9.6j^ T9.6 \hat{j} \, \text{T}

D

9.6×10−8k^ T9.6 \times 10^{-8} \hat{k} \, \text{T}

Answer

3.2×10−8k^ T3.2 \times 10^{-8} \hat{k} \, \text{T}

Explanation

Solution

For an electromagnetic wave, the electric field E⃗\vec{E} and magnetic field B⃗\vec{B} are related by the speed of light cc:
EB=c\frac{E}{B} = c
where c=3×108 m/sc = 3 \times 10^8 \, \text{m/s}.

Step 1: Calculate the magnetic field BB:
Given E=9.6 V/mE = 9.6 \, \text{V/m},
B=Ec=9.63×108=3.2×10−8 TB = \frac{E}{c} = \frac{9.6}{3 \times 10^8} = 3.2 \times 10^{-8} \, \text{T}

Step 2: Determine the direction of B⃗\vec{B}:
Since the wave travels along the X-direction and E⃗\vec{E} is along j^\hat{j}, the magnetic field B⃗\vec{B} must be perpendicular to both the direction of propagation and E⃗\vec{E}. By the right-hand rule:
B⃗ points along k^.\vec{B} \, \text{points along} \, \hat{k}.

Thus, the magnetic field at this point is 3.2×10−8k^ T.3.2 \times 10^{-8} \hat{k} \, \text{T}.

The Correct Answer is: 3.2×10−8k^ T.3.2 \times 10^{-8} \hat{k} \, \text{T}.