KCET · Physics · Motion In One Dimension
A plane electromagnetic wave of frequency \( 20 \mathrm{MHz} \) travels through a space along \( x \) direction. If the electric field vector at a certain point in space is \( 6 \mathrm{~V} \mathrm{~m}^{-1} \), what is the magnetic field vector at that point?
- A \( 2 \times 10^{-8} \mathrm{~T} \)
- B \( \frac{1}{2} \times 10^{-8} \mathrm{~T} \)
- C 2T
- D \( \frac{1}{2} T \)
Answer & Solution
Correct Answer
(A) \( 2 \times 10^{-8} \mathrm{~T} \)
Step-by-step Solution
Detailed explanation
The electric field vector and magnetic field vector are related as
\(\frac{E}{B}=c\)
We know that \( c=3 \times 10^{8} \mathrm{~ms}^{-1} \) and it is given that \( E=6 \mathrm{Vm}^{-1} \) Therefore,
\( \frac{6}{B}=3 \times 10^{8} \Rightarrow B=\frac{6}{3 \times 10^{8}}=2 \times 10^{-8} \mathrm{~T} \)
Thus, magnetic field vector is \( 2 \times 10^{-8} \mathrm{~T} \).
\(\frac{E}{B}=c\)
We know that \( c=3 \times 10^{8} \mathrm{~ms}^{-1} \) and it is given that \( E=6 \mathrm{Vm}^{-1} \) Therefore,
\( \frac{6}{B}=3 \times 10^{8} \Rightarrow B=\frac{6}{3 \times 10^{8}}=2 \times 10^{-8} \mathrm{~T} \)
Thus, magnetic field vector is \( 2 \times 10^{-8} \mathrm{~T} \).
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