JEE Main 2021

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Animated Solution for Physics - Electromagnetic Waves: A plane electromagnetic wave with frequency of $30 MHz$ travels in free space. At particular point in space and time, electric field is $6 V/m$ . The magnetic field at this point will be $x \times 1 0^{- 8} T$ . The value of $x$ is.

Q: What is the direction of the magnetic field if the electric field is along the y-axis and the wave propagates along the x-axis?

The direction of propagation is given by the cross product $\vec{E} \times \vec{B}$. If propagation is $+\hat{i}$ and $\vec{E}$ is $+\hat{j}$, then $\vec{B}$ must be along $+\hat{k}$ (z-axis) because $\hat{j} \times \hat{k} = \hat{i}$.

Enter Numerical Value:

Visualized Solution

$E$ \text{ and } $B$ \text{ Vectors}

$f = 30 MHz$
$E = 6 V/m$
$B = x \times 1 0^{- 8} T$

$B = \frac{E}{c}$

$B = \frac{E}{c}$

\text{Substitution}

$B = \frac{6}{3 \times 1 0 ^{8}}$

\text{Calculation}

$B = 2 \times 1 0^{- 8} T$

\text{Final Answer}

$x \times 1 0^{- 8} = 2 \times 1 0^{- 8}$
$x = 2$

\text{The Way Forward}

\text{In a medium: } B = \frac{E}{v}
$v = \frac{c}{n}$

00:00 / 00:00

The Sigma Insight: Characteristics of Electromagnetic Waves

Solution Diagram

Analyzing the Setup

Imagine you are standing in the middle of empty space, watching an electromagnetic wave zoom past you at the speed of light. The problem gives us a snapshot of this wave at a very specific point and time.

At this exact moment, the electric field

E

has a strength of

6 V/m

. We are also told that the frequency of this wave is

30 MHz

. Our ultimate goal is to find the strength of the magnetic field

B

at this exact same spot, which is given in the form

x \times 1 0^{- 8} T

Here is a classic trap: The frequency of

30 MHz

is completely irrelevant to finding the amplitude of the magnetic field at a specific point! It is just extra information designed to test your conceptual clarity.

The Master Equation

So, what connects the electric field, the magnetic field, and the speed of light in a vacuum?

In any electromagnetic wave traveling through free space, the electric and magnetic fields are perfectly synchronized. They reach their peaks together and hit zero together. Because of this beautiful symmetry, their magnitudes are locked in a very simple, elegant ratio.

The magnitude of the magnetic field is simply the electric field divided by the speed of light.

B = \frac{E}{c}

Final Calculation

Now, let's bring back the numbers we have. We know the electric field

E

6 V/m

, and the speed of light

c

is a universal constant,

3 \times 1 0^{8} m/s

Let's substitute these values into our master equation:

B = \frac{6}{3 \times 1 0 ^{8}}

Dividing

6

3

gives us

2

. When we bring the

1 0^{8}

from the denominator up to the numerator, the exponent changes sign, becoming

1 0^{- 8}

B = 2 \times 1 0^{- 8} T

The problem states that the magnetic field is

x \times 1 0^{- 8} T

. By comparing this expression with our calculated result, it is crystal clear that the value of

x

is exactly

2

Final Answer:

x = 2

Common Traps & FAQs

Why didn't we use the frequency of 30 MHz in our calculation?

The relationship between the magnitudes of the electric and magnetic fields at any point in free space depends only on the speed of light (

B = E / c

). The frequency determines how fast the fields oscillate, but not their relative amplitudes.

Does this formula apply if the wave is in a medium like glass or water?

No, in a dielectric medium, the speed of the wave changes to

v = c / n

, where

n

is the refractive index. The relationship then becomes

B = E / v

What is the direction of the magnetic field if the electric field is along the y-axis and the wave propagates along the x-axis?

The direction of propagation is given by the cross product

E \times B

. If propagation is

+ \hat{i}

and

E

+ \hat{j}

, then

B

must be along

+ \hat{k}

(z-axis) because

\hat{j} \times \hat{k} = \hat{i}

Are the electric and magnetic fields in phase?

Yes, in a plane electromagnetic wave in free space, the electric and magnetic fields oscillate perfectly in phase. They reach their maximum and zero values at the exact same time and position.

What does the

1 0^{- 8}

in the answer signify?

It indicates the extremely small magnitude of the magnetic field compared to the electric field. Because

c

is so large (

3 \times 1 0^{8}

B

is always much smaller than

E

when measured in standard SI units (Tesla vs. V/m).

Similar Questions

JEE Main 2019

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A plane electromagnetic wave of frequency $50 MHz$ travels in free space along the positive $x$ -direction. At a particular point in space and time, $E = 6.3 \hat{j} V/m$ . The corresponding magnetic field $B$ , at that point will be

(A)

18.9 \times 1 0^{8} \hat{k} T

(B)

6.3 \times 1 0^{- 8} \hat{k} T

(C)

18.9 \times 1 0^{- 8} \hat{k} T

(D)

2.1 \times 1 0^{- 8} \hat{k} T

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A plane electromagnetic wave of frequency $100 MHz$ is travelling in vacuum along the x-direction. At a particular point in space and time, $B = 2.0 \times 1 0^{- 8} \hat{k} T$ (where, $\hat{k}$ is unit vector along z-direction). What is $E$ at this point?

(A)

0.6 \hat{j} V/m

(B)

6.0 \hat{k} V/m

(C)

6.0 \hat{j} V/m

(D)

0.6 \hat{k} V/m

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A plane electromagnetic wave travels in free space along the x-direction. The electric field component of the wave at a particular point of space and time is $E = 6 Vm^{- 1}$ along y-direction. Its corresponding magnetic field component, $B$ would be

(A)

2 \times 1 0^{- 8} T

along z - direction

(B)

6 \times 1 0^{- 8} T

along x - direction

(C)

6 \times 1 0^{- 8} T

along z - direction

(D)

2 \times 1 0^{- 8} T

along y - direction

JEE Main 2021

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A plane electromagnetic wave of frequency $500 MHz$ is travelling in vacuum along $y$ -direction. At a particular point in space and time, $B = 8.0 \times 1 0^{- 8} \hat{z} T$ . The value of electric field at this point is (speed of light $= 3 \times 1 0^{8} ms^{- 1}$ ; $\hat{x}, \hat{y}, \hat{z}$ are unit vectors along $x, y$ and $z$ -direction).

(A)

- 24 \hat{x} V/m

(B)

2.6 \hat{x} V/m

(C)

24 \hat{x} V/m

(D)

- 2.6 \hat{x} V/m

JEE Main 2020

LEVELJEE Main

A plane electromagnetic wave of frequency $25 GHz$ is propagating in vacuum along the z-direction. At a particular point in space and time, the magnetic field is given by $B = 5 \times 1 0^{- 8} \hat{j} T$ . The corresponding electric field $E$ is (speed of light, $c = 3 \times 1 0^{8} ms^{- 1}$ )

(A)

- 1.66 \times 1 0^{- 16} \hat{i} V/m

(B)

1.66 \times 1 0^{- 16} \hat{i} V/m

(C)

- 15 \hat{i} V/m

(D)

15 \hat{i} V/m

JEE Main 2020

LEVELJEE Main

The electric field of a plane electromagnetic wave propagating along the x-direction in vacuum is $E = E_{0} \hat{j} cos (ω t - k x)$ . The magnetic field $B$ , at the moment $t = 0$ is

(A)

B = E_{0} μ_{0} ϵ_{0} cos (k x) \hat{j}

(B)

B = \frac{E _{0}}{μ _{0} ϵ _{0}} cos (k x) \hat{j}

(C)

B = \frac{E _{0}}{μ _{0} ϵ _{0}} cos (k x) \hat{k}

(D)

B = E_{0} μ_{0} ϵ_{0} cos (k x) \hat{k}

JEE Main 2020

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A plane electromagnetic wave, has frequency of $2.0 \times 1 0^{10}$ Hz and its energy density is $1.02 \times 1 0^{- 8} J/m^{3}$ in vacuum. The amplitude of the magnetic field of the wave is close to (Take, $\frac{1}{4 π ε _{0}} = 9 \times 1 0^{9} \frac{Nm ^{2}}{C ^{2}}$ and speed of light $= 3 \times 1 0^{8} ms^{- 1}$ )

If the magnetic field of a plane electromagnetic wave is given by $B = 100 \times 1 0^{- 6} sin [2 π \times 2 \times 1 0^{15} (t - \frac{x}{c})]$ then the maximum electric field associated with it is (Take, the speed of light $= 3 \times 1 0^{8}$ m/s)

(A)

6 \times 1 0^{4}

N/C

(B)

4 \times 1 0^{4}

N/C

(C)

3 \times 1 0^{4}

N/C

(D)

4.5 \times 1 0^{4}

N/C

JEE Main 2020

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For a plane electromagnetic wave, the magnetic field at a point $x$ and time $t$ is $B (x, t) = [1.2 \times 1 0^{- 7} sin (0.5 \times 1 0^{3} x + 1.5 \times 1 0^{11} t) \hat{k}] T$ . The instantaneous electric field $E$ corresponding to $B$ is (Given, speed of light, $c = 3 \times 1 0^{8} m/s$ )

(A)

E (x, t) = [- 36 sin (0.5 \times 1 0^{3} x + 1.5 \times 1 0^{11} t) \hat{j}] \frac{V}{m}

(B)

E (x, t) = [36 sin (1 \times 1 0^{3} x + 0.5 \times 1 0^{11} t) \hat{j}] \frac{V}{m}

(C)

E (x, t) = [36 sin (0.5 \times 1 0^{3} x + 1.5 \times 1 0^{11} t) \hat{k}] \frac{V}{m}

(D)

E (x, t) = [- 36 sin (1 \times 1 0^{3} x + 1.5 \times 1 0^{11} t) \hat{j}] \frac{V}{m}

JEE Main 2020

LEVELJEE Main

The magnetic field of a plane electromagnetic wave is $B = 3 \times 1 0^{- 8} sin [200 π (y + c t)] \hat{i}$ T, where $c = 3 \times 1 0^{8}$ ms $^{- 1}$ is the speed of light. The corresponding electric field is

(A)

E = 9 sin [200 π (y + c t)] \hat{k}

V/m

(B)

E = - 1 0^{- 6} sin [200 π (y + c t)] \hat{k}

V/m

(C)

E = 3 \times 1 0^{- 8} sin [200 π (y + c t)] \hat{k}

V/m

(D)

E = - 9 sin [200 π (y + c t)] \hat{k}

V/m

Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E \text{ and } .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B \text{ Vectors}

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B=cE​

\text{Substitution}

\text{Calculation}

\text{Final Answer}

\text{The Way Forward}

The Sigma Insight: Characteristics of Electromagnetic Waves

Analyzing the Setup

The Master Equation

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The electric field of a plane electromagnetic wave propagating along the x-direction in vacuum is E=E0​j^​cos(ωt−kx). The magnetic field B, at the moment t=0 is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The magnetic field of a plane electromagnetic wave is B=3×10−8sin[200π(y+ct)]i^ T, where c=3×108 ms−1 is the speed of light. The corresponding electric field is

$E$ \text{ and } $B$ \text{ Vectors}

$B = \frac{E}{c}$

The electric field of a plane electromagnetic wave propagating along the x-direction in vacuum is $E = E_{0} \hat{j} cos (ω t - k x)$ . The magnetic field $B$ , at the moment $t = 0$ is

The magnetic field of a plane electromagnetic wave is $B = 3 \times 1 0^{- 8} sin [200 π (y + c t)] \hat{i}$ T, where $c = 3 \times 1 0^{8}$ ms $^{- 1}$ is the speed of light. The corresponding electric field is