The dimension of mu0epsilon0 is equal to that of: (mu0= Vacuum permeability and epsilon0= Vacuum permittivity) [cite…

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Q jee_main_2025_28_jan_evening Electric and Magnetic Field Vectors

The magnetic field of an E.M. wave is given by

B=( \sqrt(3)2 i+(1)/(2) j)30~ [ω(t-(z)/(c))]

(S.I. Units). The corresponding electric field in S.I. units is:

A. $E = ((1)/(2) i - \sqrt(3)2 j)30c [ω (t - zc)]$
B. \vec{E} = \left(\frac{3}{4}\hat{i} + \frac{1}{4}\hat{j}\right) 30c \cos \left[\omega \left(t - \frac{z}{c}\right)\right]
C. $E = ((1)/(2) i + \sqrt(3)2 j)30c [ω (t + zc)]$
D. $E = ( \sqrt(3)2 i -(1)/(2) j)30c [ω (t + zc)]$

Solution

Related Formula

For a plane electromagnetic wave propagating in a given direction:

Peak electric field amplitude relates to peak magnetic field amplitude via:

E₀ = B₀ \cdot c

The directional orientation unit vectors satisfy the cross product relation:

E = B \times c

where $c$ $\hat{c}$ points along the wave propagation vector direction.

Core Logic

Given the wave equation format, the phase term $(t - (z)/(c))$ $\left(t - \frac{z}{c}\right)$ shows that propagation is along the positive z-axis :

c = k

The magnetic field direction unit vector is :

B = \sqrt(3)2 i + (1)/(2) j

Compute the electric field direction vector using the cross product relation :

E = B \times k = ( \sqrt(3)2 i + (1)/(2) j) \times k

E = \sqrt(3)2( i \times k) + (1)/(2)( j \times k)

Using unit vector properties ( $i \times k = - j$ $\hat{i} \times \hat{k} = -\hat{j}$ and $j \times k = i$ $\hat{j} \times \hat{k} = \hat{i}$):

E = - \sqrt(3)2 j + (1)/(2) i = (1)/(2) i - \sqrt(3)2 j

With peak amplitude $E₀ = 30c$ $E_0 = 30c$ , the resulting vector equation is:

E = ((1)/(2) i - \sqrt(3)2 j)30c [ω(t-(z)/(c))]

Pattern Recognition

The vectors $E$ $\vec{E}$, $B$ $\vec{B}$, and the propagation direction are always mutually perpendicular. Since $E \cdot B = 0$ $\vec{E} \cdot \vec{B} = 0$, you can quickly double-check your answer by verifying that the \dot product of the final $E$ $\vec{E}$ and $B$ $\vec{B}$ direction options equals zero.

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Class 12 Physics: Electromagnetic Waves

Q jee_main_2025_29_jan_morning Properties of EM Waves

Given below are two statements : one is labelled as Assertion (A) and other is labelled as Reason (R). Assertion (A) : Electromagnetic waves carry energy but not momentum. Reason (R): Mass of a photon is zero. In the light of the above statements, choose the most appropriate answer from the options given below:

A. (A) is true but (R) is false.
B. (A) is false but (R) is true.
C. Both (A) and (R) are true but (R) is not the correct explanation of (A).
D. Both (A) and (R) are true and (R) is the correct explanation of (A).

Solution

Related Formula

p = (E)/(c)

Core Logic

Assertion (A) is false because electromagnetic waves carry both energy and finite radiation momentum ( $p = E/c$ $p = E/c$). Reason (R) is correct because the rest mass of a photon equals zero.

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Class 12 Physics: Electromagnetic Waves

Q38 jee_main_2024_01_february_morning Displacement Current

A parallel plate capacitor has a capacitance

C = 200~pF

. It is connected to

230~V

ac supply with an angular frequency

300~rad/s

. The rms value of conduction current in the circuit and displacement current in the capacitor respectively are:

A. $1.38~μ A and 1.38~μ A$
B. $14.3~μ A and 143~μ A$
C. $13.8~μ A and 138~μ A$
D. $13.8~μ A and 13.8~μ A$

Solution

Related Formula

Capacitive reactance:

XC = (1)/(ω C)

Conduction Current ( $Irms$ $I_{\text{rms}}$):

Irms = VrmsXC = Vrms \cdot ω C

Continuity relation: $Ic = Id$ $I_c = I_d$

Core Logic

Given parameters: $C = 200~pF = 200 \times 10⁻¹²~F$ , $Vrms = 230~V$ $V_{\text{rms}} = 230\mathrm{~V}$, $ω = 300~rad/s$ $\omega = 300\mathrm{~rad/s}$.

Calculate current:

I = Vrms \cdot ω C = 230 \times 300 \times 200 \times 10⁻¹²

Step 1: Simplify Numerical Calculation

$I = 230 \times 60000 \times 10⁻¹² = 13.8 \times 10⁻⁶~A = 13.8~μ A$

By Maxwell's electromagnetic formulation, the displacement current $Id$ $I_d$ within the dielectric space matches the exterior conduction current $Ic$ $I_c$ seamlessly in magnitude.

Pattern Recognition

Factual invariant: Conduction current always equals displacement current inside a standard layout loop context ( $Ic = Id$ $I_c = I_d$).

Chapter Mix

Class 12 Physics: Electromagnetic Waves Class 12 Physics: Alternating Current

Q45 jee_main_2024_29_january_evening Electric and Magnetic Field Relations

A plane electromagnetic wave of frequency

35 MHz

travels in free space along the

X

-direction. At a particular point (in space and time)

E = 9.6 j V/m

. The value of magnetic field at this point is:

A. $3.2 \times 10⁻⁸ k T$ $3.2 \times 10^{-8}\hat{k}\text{ T}$
B. $3.2 \times 10⁻⁸ i T$ $3.2 \times 10^{-8}\hat{i}\text{ T}$
C. $9.6 j T$ $9.6\hat{j}\text{ T}$
D. $9.6 \times 10⁻⁸ k T$ $9.6 \times 10^{-8}\hat{k}\text{ T}$

Solution

Related Formula

The relation between the amplitudes of electric field $E$ $E$ and magnetic field $B$ $B$ in an EM wave is:

c = (E)/(B)

where:

$c = 3 \times 10⁸ m/s$ $c = 3 \times 10^8\text{ m/s}$ is the speed of light.

The directions are related by the vector cross product:

E \times B = v

where $v$ $\hat{v}$ is the direction of propagation of the EM wave.

Core Logic

Given:

Wave propagation direction, $v = i$ $\hat{v} = \hat{i}$ (along X-direction)
Electric field vector direction, $E = j$ $\hat{E} = \hat{j}$
Electric field magnitude, $E = 9.6 V/m$ $E = 9.6\text{ V/m}$

Step 1: Calculate Magnetic Field Magnitude

Using the magnitude relation:

B = (E)/(c) = (9.6)/(3 \times 10⁸) = 3.2 \times 10⁻⁸ T

Step 2: Determine Magnetic Field Direction

Using the direction cross-product rule:

E \times B = v

j \times B = i

Since we know that $j \times k = i$ $\hat{j} \times \hat{k} = \hat{i}$, the direction of the magnetic field must be $k$ $\hat{k}$:

B = k

Combining the magnitude and direction:

B = 3.2 \times 10⁻⁸ k T

Pattern Recognition

A propagation along $+x$ $+x$ with electric field along $+y$ $+y$ strictly mandates the magnetic field must point along $+z$ $+z$ ( $+y \times +z = +x$ $+y \times +z = +x$). This instantly eliminates Option 2 and Option 3, leaving only magnitude verification.

Chapter Mix

Class 12 Physics: Electromagnetic Waves

Q46 jee_main_2024_27_jan_morning Energy Density and Intensity

A plane electromagnetic wave propagating in x-direction is described by

Ey = (200 Vm⁻¹) [1.5 \times 10⁷t - 0.05x]

. The intensity of the wave is (Use

ε₀ = 8.85 \times 10⁻¹² C²N⁻¹m⁻²

):

A. $35.4 Wm⁻²$ $35.4\text{ Wm}^{-2}$
B. $53.1 Wm⁻²$ $53.1\text{ Wm}^{-2}$
C. $26.6 Wm⁻²$ $26.6\text{ Wm}^{-2}$
D. $106.2 Wm⁻²$ $106.2\text{ Wm}^{-2}$

Solution

Related Formula

I = (1)/(2) ε₀ E₀² c

Where $E₀$ $E_0$ is the amplitude of the electric field ( $200 V/m$ $200\text{ V/m}$) and $c = 3 \times 10⁸ m/s$ $c = 3 \times 10^8\text{ m/s}$.

Core Logic

Substitute the constants into the equation:

I = (1)/(2) \times (8.85 \times 10⁻¹²) \times (200)² \times (3 \times 10⁸)

Step 1: Compute value

I = (1)/(2) \times 8.85 \times 10⁻¹² \times 4 \times 10⁴ \times 3 \times 10⁸

I = 2 \times 8.85 \times 3 \times 10⁰

I = 53.1 W/m²

Pattern Recognition

Isolate powers of ten first ( $10⁻¹² \times 10⁴ \times 10⁸ = 10⁰$ ) to streamline intermediate tracking accuracy on calculation variables.

Chapter Mix

Class 12 Physics: Electromagnetic Waves

The dimension of μ₀ε₀$\frac{\mu_{0}}{\epsilon_{0}}$ is equal to that of: (μ₀=$\mu_{0}=$ Vacuum permeability and ε₀=$\epsilon_{0}=$ Vacuum permittivity) [cite: 33, 34]

Solution & Explanation

Related Formula

Core Logic

Pattern Recognition

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Reference Study Guides

More Electromagnetic Waves Previous-Year Questions — Page 5

Solution

Related Formula

Core Logic

Pattern Recognition

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Solution

Related Formula

Core Logic

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Solution

Related Formula

Core Logic

Step 1: Simplify Numerical Calculation

Pattern Recognition

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Solution

Related Formula

Core Logic

Step 1: Calculate Magnetic Field Magnitude

Step 2: Determine Magnetic Field Direction

Pattern Recognition

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Solution

Related Formula

Core Logic

Step 1: Compute value

Pattern Recognition

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The dimension of $μ₀ε₀$ $\frac{\mu_{0}}{\epsilon_{0}}$ is equal to that of: ( $μ₀=$ $\mu_{0}=$ Vacuum permeability and $ε₀=$ $\epsilon_{0}=$ Vacuum permittivity) [cite: 33, 34]