Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
An electromagnetic wave of frequency $45 \mathrm{MHz}$ travels in free space along $X$-axis. At some point and at some instant, the electric field has a maximum value of $750 \mathrm{NC}^{-1}$ along $Y$-axis. The magnetic field at this position and time is
PhysicsElectromagnetic WavesAP EAMCETAP EAMCET 2018 (24 Apr Shift 1)
Options:
  • A $2.5 \times 10^{-6} \hat{\mathbf{j}}$
  • B $5 \times 10^{-6} \hat{\mathbf{k}} \mathrm{T}$
  • C $2.5 \times 10^{-6} \hat{\mathbf{k}} \mathrm{T}$
  • D $2.5 \times 10^{-6} \hat{\mathbf{i}} \mathrm{T}$
Solution:
2477 Upvotes Verified Answer
The correct answer is: $2.5 \times 10^{-6} \hat{\mathbf{k}} \mathrm{T}$
Relation between electric and magnetic field in electromagnetic wave
$\begin{aligned}
& B=\frac{E}{c} \\
& E=750 \mathrm{~N} / \mathrm{C}
\end{aligned}$
Speed of light,
$\begin{aligned}
c & =3 \times 10^8 \mathrm{~m} / \mathrm{s} \\
B & =\frac{750}{3 \times 10^8} \\
B & =2.5 \times 10^{-6} \mathrm{~T}
\end{aligned}$
As, $B$ must be perpendicular to both $\mathbf{c}$ and $\mathbf{E}$, i.e.it is along $Z$-axis.
$B=2.5 \times 10^{-6} \hat{\mathbf{k}} \mathrm{T}$

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.