Match List I with List II:
List-I List-II
A. Gauss's law of magnetostatics I. $oint vecE cdot mathrmdveca = frac1varepsilon_0 int rho mathrmdV$
B. Faraday's law of electro magnetic induction II. $oint vecB cdot mathrmdveca = 0$
C. Ampere's law III. $oint vecE cdot mathrmdvecl = -fracmathrmdmathrmdtint vecB cdot mathrmdveca$
D. Gauss's law of electrostatics IV. $oint vecB cdot mathrmdvecl = mu_0 I$
Choose the correct answer from the options given below:

List-I	List-II
A. Gauss's law of magnetostatics	I. $oint vecE cdot mathrmdveca = frac1varepsilon_0 int rho mathrmdV$
B. Faraday's law of electro magnetic induction	II. $oint vecB cdot mathrmdveca = 0$
C. Ampere's law	III. $oint vecE cdot mathrmdvecl = -fracmathrmdmathrmdtint vecB cdot mathrmdveca$
D. Gauss's law of electrostatics	IV. $oint vecB cdot mathrmdvecl = mu_0 I$

A. $textA-I, B-III, C-IV, D-II$ $\text{A-I, B-III, C-IV, D-II}$
B. $textA-III, B-IV, C-I, D-II$ $\text{A-III, B-IV, C-I, D-II}$
C. $textA-IV, B-II, C-III, D-I$ $\text{A-IV, B-II, C-III, D-I}$
D. $textA-II, B-III, C-IV, D-I$ $\text{A-II, B-III, C-IV, D-I}$

Solution & Explanation

### Core Logic Match each law with its corresponding mathematical expression (Maxwell's equations). (A) Gauss's law of magnetostatics: The net magnetic flux through any closed surface is zero.

oint vecB cdot mathrmdveca = 0

(Matches II). (B) Faraday's law of electromagnetic induction: The induced electromotive force in any closed circuit is equal to the negative of the time rate of change of the magnetic flux.

oint vecE cdot mathrmdvecl = -fracmathrmdmathrmdt int vecB cdot mathrmdveca

(Matches III). (C) Ampere's law: The line integral of the magnetic field around a closed loop is proportional to the electric current passing through the loop.

oint vecB cdot mathrmdvecl = mu_0 I

(Matches IV). (D) Gauss's law of electrostatics: The electric flux through any closed surface is proportional to the enclosed electric charge.

oint vecE cdot mathrmdveca = frac1varepsilon_0 int rho mathrmdV

(Matches I). ### Step 1: Final Match A

rightarrow

II B

rightarrow

III C

rightarrow

IV D

rightarrow

I This matches option (4). ### Pattern Recognition These are the fundamental Maxwell equations in integral form. Memorizing their direct mappings guarantees quick marks. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electromagnetic Waves Class 12 Physics: Electromagnetic Induction

Reference Study Guides

Electromagnetic Waves

More Electromagnetic Waves Previous-Year Questions — Page 3

Q19 jee_main_2025_28_jan_evening Electric and Magnetic Field Vectors

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

vecB=left(fracsqrt32hati+frac12hatjright)30~sinleft[omegaleft(t-fraczcright)right]

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

A. $vecmathrmE = left(frac12hatmathrmi -fracsqrt32hatmathrmjright)30mathrmcsin left[omega left(mathrmt - fracmathrmzmathrmcright)right]$
B. $vecmathrmE = left(frac34\i +frac14hatmathrmjright)30mathrmccos left[omega left(mathrmt - fracmathrmzmathrmcright)right]$
C. $vecmathrmE = left(frac12hatmathrmi +fracsqrt32hatmathrmjright)30mathrmcsin left[omega left(mathrmt + fracmathrmzmathrmcright)right]$
D. $vecmathrmE = left(fracsqrt32hatmathrmi -frac12hatmathrmjright)30mathrmcsin left[omega left(mathrmt + fracmathrmzmathrmcright)right]$

Solution

### Related Formula For a plane electromagnetic wave propagating in a given direction: 1. Peak electric field amplitude relates to peak magnetic field amplitude via:

E_0 = B_0 cdot c

2. The directional orientation unit vectors satisfy the cross product relation:

hatE = hatB times hatc

where

hatc

points along the wave propagation vector direction. ### Core Logic Given the wave equation format, the phase term

left(t - fraczcright)

shows that propagation is along the positive z-axis :

hatc = hatk

The magnetic field direction unit vector is :

hatB = fracsqrt32hati + frac12hatj

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

hatE = hatB times hatk = left(fracsqrt32hati + frac12hatjright) times hatk

hatE = fracsqrt32(hati times hatk) + frac12(hatj times hatk)

Using unit vector properties (

hati times hatk = -hatj

and

hatj times hatk = hati

hatE = -fracsqrt32hatj + frac12hati = frac12hati - fracsqrt32hatj quad text

With peak amplitude

E_0 = 30c

, the resulting vector equation is:

vecE = left(frac12hati - fracsqrt32hatjright)30csinleft[omegaleft(t-fraczcright)right]

### Pattern Recognition The vectors

vecE

vecB

, and the propagation direction are always mutually perpendicular. Since

vecE cdot vecB = 0

, you can quickly double-check your answer by verifying that the \dot product of the final

vecE

and

vecB

direction options equals zero. ### Evaluation Rubric / Model Answer null ### Chapter Mix 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 = fracEc

### Core Logic Assertion (A) is false because electromagnetic waves carry both energy and finite radiation momentum (

p = E/c

). Reason (R) is correct because the rest mass of a photon equals zero. ### Chapter Mix Class 12 Physics: Electromagnetic Waves

Q38 jee_main_2024_01_february_morning Displacement Current

A parallel plate capacitor has a capacitance

C = 200mathrm~pF

. It is connected to

230mathrm~V

ac supply with an angular frequency

300mathrm~rad/s

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

A. $1.38mathrm~mu Atext and 1.38mathrm~mu A$
B. $14.3mathrm~mu Atext and 143mathrm~mu A$
C. $13.8mathrm~mu Atext and 138mathrm~mu A$
D. $13.8mathrm~mu Atext and 13.8mathrm~mu A$

Solution

### Related Formula Capacitive reactance:

X_C = frac1omega C

Conduction Current (

I_textrms

I_textrms = fracV_textrmsX_C = V_textrms cdot omega C

Continuity relation:

I_c = I_d

### Core Logic Given parameters:

C = 200mathrm~pF = 200 times 10^-12mathrm~F

V_textrms = 230mathrm~V

omega = 300mathrm~rad/s

. Calculate current:

I = V_textrms cdot omega 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^-6mathrm~A = 13.8mathrm~mu A

By Maxwell's electromagnetic formulation, the displacement current

I_d

within the dielectric space matches the exterior conduction current

I_c

seamlessly in magnitude. ### Pattern Recognition Factual invariant: Conduction current always equals displacement current inside a standard layout loop context (

I_c = I_d

). ### Evaluation Rubric / Model Answer null ### 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

35text MHz

travels in free space along the

X

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

vecE = 9.6hatjtext V/m

. The value of magnetic field at this point is:

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

Solution

### Related Formula The relation between the amplitudes of electric field

E

and magnetic field

B

in an EM wave is:

c = fracEB

where: *

c = 3 times 10^8text m/s

is the speed of light. The directions are related by the vector cross product:

hatE times hatB = hatv

where

hatv

is the direction of propagation of the EM wave. ### Core Logic Given: * Wave propagation direction,

hatv = hati

(along X-direction) * Electric field vector direction,

hatE = hatj

* Electric field magnitude,

E = 9.6text V/m

### Step 1: Calculate Magnetic Field Magnitude Using the magnitude relation:

B = fracEc = frac9.63 times 10^8 = 3.2 times 10^-8text T

### Step 2: Determine Magnetic Field Direction Using the direction cross-product rule:

hatE times hatB = hatv

hatj times hatB = hati

Since we know that

hatj times hatk = hati

, the direction of the magnetic field must be

hatk

hatB = hatk

Combining the magnitude and direction:

vecB = 3.2 times 10^-8hatktext T

### Pattern Recognition A propagation along

+x

with electric field along

+y

strictly mandates the magnetic field must point along

+z

(

+y times +z = +x

). This instantly eliminates Option 2 and Option 3, leaving only magnitude verification. ### Evaluation Rubric / Model Answer null ### 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

E_y = (200text Vm^-1)sin[1.5 times 10^7t - 0.05x]

. The intensity of the wave is (Use

epsilon_0 = 8.85 times 10^-12text C^2textN^-1textm^-2

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

Solution

### Related Formula

I = frac12 epsilon_0 E_0^2 c

Where

E_0

is the amplitude of the electric field (

200text V/m

) and

c = 3 times 10^8text m/s

. ### Core Logic Substitute the constants into the equation:

I = frac12 times (8.85 times 10^-12) times (200)^2 times (3 times 10^8)

### Step 1: Compute value

I = frac12 times 8.85 times 10^-12 times 4 times 10^4 times 3 times 10^8

I = 2 times 8.85 times 3 times 10^0

I = 53.1text W/m^2

### Pattern Recognition Isolate powers of ten first (

10^-12 times 10^4 times 10^8 = 10^0

) to streamline intermediate tracking accuracy on calculation variables. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electromagnetic Waves

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