JEE Main 2015 — Physics Electromagnetism

2015mcqmedium

An electromagnetic wave travelling in the $x-$ direction has frequency of $2\times {10}^{14}Hz$ and electric field amplitude of $27V{m}^{–1}$ oscillates in $Y-$ direction. From the options given below, which one describes the magnetic field for this wave?

Official previous-year question

Held on 10 Apr 2015 · Verified 6 Jul 2026.

Options

A
$\vec{B}(x, t)=(9\times {10}^{-8}T)\hat{j}$ $\mathrm{sin}[1.5\times {10}^{-6} x-2\times {10}^{14}t]$
B
$\vec{B}(x, t)=(9\times {10}^{-8}T)\hat{i}\mathrm{sin}[2\pi (1.5\times {10}^{-8} x-2\times {10}^{14}t)]$
C
$\vec{B}$ $(x,t)=(3\times {10}^{-8}T)\hat{j}\mathrm{sin}2\pi [(\frac{x}{1.5\times {10}^{-8}})-2\times {10}^{14}t]$
D
$\vec{B}$ $(x,t)=(9\times {10}^{-8}T)\hat{k}sin2\pi [(\frac{x}{1.5\times {10}^{-6}})-2\times {10}^{14}t]$

Did you get this right?

Solution

When undefined

Then $B = {B}_{0}\mathrm{sin}⁡$ $( kx-\omega t )$

Of light in travelling along $\hat{i}$ then $\vec{B}$ in either along $\hat{j}$ or $\hat{k}$ .

$\therefore$ speed of light $C= \frac{{E}_{0}}{{B}_{0}} \Rightarrow {B}_{0}= \frac{{E}_{0}}{C}$

$\Rightarrow {B}_{0}= \frac{27}{3\times {10}^{8}}=9 \times {10}^{-8} T$

also, $\omega =2\pi f=2\pi \times 2 \times {10}^{14}=4 \pi \times {10}^{14}$

$k=\frac{\omega }{c} =\frac{ 4\pi \times { 10}^{ 14} }{ 3\times { 10}^{8} }$

$=( \frac{ 2\pi }{ 1.5\times { 10}^{ -6} } )$

Looking into the option the correct

Answer is $\vec{B}=9\times {10}^{-8}\mathrm{sin}⁡2\pi (\frac{x}{1.5\times {10}^{-6}}-2\times {10}^{14}t)\hat{k}$

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Options

$\vec{B}(x, t)=(9\times {10}^{-8}T)\hat{j}$ $\mathrm{sin}[1.5\times {10}^{-6} x-2\times {10}^{14}t]$

$\vec{B}(x, t)=(9\times {10}^{-8}T)\hat{i}\mathrm{sin}[2\pi (1.5\times {10}^{-8} x-2\times {10}^{14}t)]$

$\vec{B}$ $(x,t)=(3\times {10}^{-8}T)\hat{j}\mathrm{sin}2\pi [(\frac{x}{1.5\times {10}^{-8}})-2\times {10}^{14}t]$

$\vec{B}$ $(x,t)=(9\times {10}^{-8}T)\hat{k}sin2\pi [(\frac{x}{1.5\times {10}^{-6}})-2\times {10}^{14}t]$

Solution

When undefined

Then $B = {B}_{0}\mathrm{sin}⁡$ $( kx-\omega t )$

Of light in travelling along $\hat{i}$ then $\vec{B}$ in either along $\hat{j}$ or $\hat{k}$ .

$\therefore$ speed of light $C= \frac{{E}_{0}}{{B}_{0}} \Rightarrow {B}_{0}= \frac{{E}_{0}}{C}$

$\Rightarrow {B}_{0}= \frac{27}{3\times {10}^{8}}=9 \times {10}^{-8} T$

also, $\omega =2\pi f=2\pi \times 2 \times {10}^{14}=4 \pi \times {10}^{14}$

$k=\frac{\omega }{c} =\frac{ 4\pi \times { 10}^{ 14} }{ 3\times { 10}^{8} }$

$=( \frac{ 2\pi }{ 1.5\times { 10}^{ -6} } )$

Looking into the option the correct

Answer is $\vec{B}=9\times {10}^{-8}\mathrm{sin}⁡2\pi (\frac{x}{1.5\times {10}^{-6}}-2\times {10}^{14}t)\hat{k}$

JEE Main 2015 — Physics Electromagnetism

Options

Did you get this right?

Solution

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JEE Main 2015 — Physics Electromagnetism

Options

Did you get this right?

Solution

Your note

JEE Main Electromagnetism in other years

More JEE Main Electromagnetism questions

Other Physics topics for JEE Main

Mechanics

Modern Physics

Optics

Thermodynamics

Waves & Oscillations

Keep practicing Electromagnetism