Q155 / 160 · AP EAMCET 2018 (24 Apr Shift 1) · Physics
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
- A$2.5 \times 10^{-6} \hat{\mathbf{j}}$
- B$5 \times 10^{-6} \hat{\mathbf{k}} \mathrm{T}$
- Correct answerC$2.5 \times 10^{-6} \hat{\mathbf{k}} \mathrm{T}$
- D$2.5 \times 10^{-6} \hat{\mathbf{i}} \mathrm{T}$
Solution
Relation between electric and magnetic field in electromagnetic wave
$\begin{aligned}<br />& B=\frac{E}{c} \\<br />& E=750 \mathrm{~N} / \mathrm{C}<br />\end{aligned}$
Speed of light,
$\begin{aligned}<br />c & =3 \times 10^8 \mathrm{~m} / \mathrm{s} \\<br />B & =\frac{750}{3 \times 10^8} \\<br />B & =2.5 \times 10^{-6} \mathrm{~T}<br />\end{aligned}$
As, $B$ must be perpendicular to both $\mathbf{c}$ and $\mathbf{E}$, i.e.it is along $Z$-axis.
$B=2.5 \times 10^{-6} \hat{\mathbf{k}} \mathrm{T}$
$\begin{aligned}<br />& B=\frac{E}{c} \\<br />& E=750 \mathrm{~N} / \mathrm{C}<br />\end{aligned}$
Speed of light,
$\begin{aligned}<br />c & =3 \times 10^8 \mathrm{~m} / \mathrm{s} \\<br />B & =\frac{750}{3 \times 10^8} \\<br />B & =2.5 \times 10^{-6} \mathrm{~T}<br />\end{aligned}$
As, $B$ must be perpendicular to both $\mathbf{c}$ and $\mathbf{E}$, i.e.it is along $Z$-axis.
$B=2.5 \times 10^{-6} \hat{\mathbf{k}} \mathrm{T}$
Hence, the correct answer is option: 'C'.