Match List I with List II: List I List II A. B cdot dl = mu0 ic + mu0 0 (dphiE/dt) I. Gauss' law for electricity B. E cdot dl = -(dphiB/dt) II. Gauss' law for magnetism C. E cdot dA = (Q/0) III. Faraday law D. B cdot dA = 0 IV. Ampere - Maxwell law Choose the correct answer from the options given below:

Let's review the fundamental Maxwell's equations: 1. **Ampere - Maxwell Law** relates the magnetic path integral to conduction current and displacement current: 2. **Faraday's Law of Induction** states that changing magnetic flux induces an electromotive force (EMF): 3. **Gauss's Law for Electricity** relates net electric flux to enclosed charge: 4. **Gauss's Law for Magnetism** states that magnetic monopoles do not exist: The match configurations are: * A arrow IV * B arrow III * C arrow I * D arrow II This perfectly corresponds to Option (3). Understand the integral geometries: Path integrals (line integrals cdot dl) correspond to circulating fields (induction laws like Ampere/Faraday). Surface integrals (flux integrals cdot dA) correspond to bounded charge states (Gauss laws). Class 12 Physics: Electromagnetic Waves Class 12 Physics: Electrostatics Class 12 Physics: Magnetism and Matter

Match List I with List II: List I List II A. B cdot dl = mu0 ic + mu0 0 (dphiE/dt) I. Gauss' law for electricity B. E cdot dl = -(dphiB/dt) II. Gauss' law for magnetism C. E cdot dA = (Q/0) III. Farad

Let's review the fundamental Maxwell's equations: 1. **Ampere - Maxwell Law** relates the magnetic path integral to conduction current and displacement current: 2. **Faraday's Law of Induction** states that changing magnetic flux induces an electromotive force (EMF): 3. **Gauss's Law for Electricity** relates net electric flux to enclosed charge: 4. **Gauss's Law for Magnetism** states that magnetic monopoles do not exist: The match configurations are: * A arrow IV * B arrow III * C arrow I * D

Match List I with List II: List I List II A. B cdot dl = mu0 ic + mu0 0 (dphiE/dt) I. Gauss' law for electricity B. E cdot dl = -(dphiB/dt) II. Gauss' law for magnetism C. E cdot dA = (Q/0) III. Faraday law D. B cdot dA = 0 IV. Ampere - Maxwell law Choose the correct answer from the options given below:

Let's review the fundamental Maxwell's equations: 1. **Ampere - Maxwell Law** relates the magnetic path integral to conduction current and displacement current: 2. **Faraday's Law of Induction** states that changing magnetic flux induces an electromotive force (EMF): 3. **Gauss's Law for Electricity** relates net electric flux to enclosed charge: 4. **Gauss's Law for Magnetism** states that magnetic monopoles do not exist: The match configurations are: * A arrow IV * B arrow III * C arrow I * D arrow II This perfectly corresponds to Option (3). Understand the integral geometries: Path integrals (line integrals cdot dl) correspond to circulating fields (induction laws like Ampere/Faraday). Surface integrals (flux integrals cdot dA) correspond to bounded charge states (Gauss laws). Class 12 Physics: Electromagnetic Waves Class 12 Physics: Electrostatics Class 12 Physics: Magnetism and Matter

Match List I with List II: List I List II A. B cdot dl = mu0 ic + mu0 0 (dphiE/dt) I. Gauss' law for electricity B. E cdot dl = -(dphiB/dt) II. Gauss' law for magnetism C. E cdot dA = (Q/0) III. Farad

Let's review the fundamental Maxwell's equations: 1. **Ampere - Maxwell Law** relates the magnetic path integral to conduction current and displacement current: 2. **Faraday's Law of Induction** states that changing magnetic flux induces an electromotive force (EMF): 3. **Gauss's Law for Electricity** relates net electric flux to enclosed charge: 4. **Gauss's Law for Magnetism** states that magnetic monopoles do not exist: The match configurations are: * A arrow IV * B arrow III * C arrow I * D

Match List I with List II: List I List II A. B cdot dl = mu0 ic + mu0 0 (dphiE/dt) I. Gauss' law for electricity B. E…

List I	List II
A. $oint vecB cdot dvecl = mu_0 i_c + mu_0 varepsilon_0 fracdphi_Edt$	I. Gauss' law for electricity
B. $oint vecE cdot dvecl = -fracdphi_Bdt$	II. Gauss' law for magnetism
C. $oint vecE cdot dvecA = fracQvarepsilon_0$	III. Faraday law
D. $oint vecB cdot dvecA = 0$ $\oint \vec{B} \cdot d\vec{A} = 0$	IV. Ampere - Maxwell law

Q35 jee_main_2026_21_jan_morning Electric and Magnetic Fields

The electric field a plane electromagnetic wave is given by :

E_y = 69 sin [ 0.6 times 10^3 x - 1.8 times 10^11 t ]text V/m.

The expression for magnetic field associated with this electromagnetic wave is ____ T.

A. $mathrmB_z = 2.3times 10^-7sin [0.6times 10^3mathrmx - 1.8times 10^11mathrmt]$
B. $mathrmB_z = 2.3times 10^-7sin [0.6times 10^3mathrmx + 1.8times 10^11mathrmt]$
C. $mathrmB_y = 69sin [0.6times 10^3mathrmx + 1.8times 10^11mathrmt]$
D. $mathrmB_y = 2.3times 10^-7sin [0.6times 10^3mathrmx - 1.8times 10^11mathrmt]$

Solution

### Related Formula

B_0 = fracE_0c

hatc = hatE times hatB

### Core Logic The phase of the wave is

(0.6 times 10^3 x - 1.8 times 10^11 t)

. This indicates the wave propagates in the

+x

direction, so

hatc = hati

. The electric field oscillates along the

y

-axis, so

hatE = hatj

. From

hatB = hatc times hatE

, we have

hatB = hati times hatj = hatk

. So, the magnetic field is along the

z

-axis (

B_z

). ### Step 1: Calculate Amplitude of B Wave speed

v = c = fracomegak = frac1.8 times 10^110.6 times 10^3 = 3 times 10^8text m/s

. The amplitude of the magnetic field is:

B_0 = fracE_0c = frac693 times 10^8 = 23 times 10^-8 = 2.3 times 10^-7text T

The phase remains exactly the same as the electric field:

B_z = 2.3 times 10^-7 sin(0.6 times 10^3 x - 1.8 times 10^11 t)

### Pattern Recognition

B_0 = E_0/c

gives the magnitude. The vector identity

hatB = hatv times hatE

gives the direction. Phase part never changes sign or terms between E and B equations. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electromagnetic Waves

Q18 jee_main_2025_02_april_evening Speed of Electromagnetic Waves

If

mu_0

and

varepsilon_0

are the permeability and permittivity of free space, respectively, then the dimension of

left(frac1mu_0varepsilon_0right)

is:

A. $mathrmL / mathrmT^2$ $\mathrm{L} / \mathrm{T}^{2}$
B. $mathrmL^2 / mathrmT^2$ $\mathrm{L}^2 / \mathrm{T}^2$
C. $mathrmT^2 /mathrmL$ $\mathrm{T}^2 /\mathrm{L}$
D. $mathrmT^2 /mathrmL^2$ $\mathrm{T}^2 /\mathrm{L}^2$

Solution

### Related Formula Maxwell's relation for the speed of light (

c

) in vacuum:

c = frac1sqrtmu_0 varepsilon_0

### Core Logic Square both sides of the speed of light formula:

c^2 = frac1mu_0 varepsilon_0

Since

c

represents the speed of light (velocity), its dimensional formula is:

[c] = [L T^-1]

Therefore, the dimensions of

c^2

are:

[c^2] = [L T^-1]^2 = [L^2 T^-2]

### Step 1: Match options Express the dimensions in terms of the given variable ratios:

[c^2] = fracL^2T^2

This perfectly matches Option (2). ### Pattern Recognition Sees: Dimensions of vacuum permittivity-permeability inverse product. Trap: Trying to find the separate dimensions of both

mu_0

and

varepsilon_0

, then conducting manual division. This is extremely slow and prone to algebraic error. Shortcut: Directly identify the term as

c^2

. The velocity squared has units of

mathrmm^2mathrms^-2

, giving dimensions of

mathrmL^2mathrmT^-2

immediately. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electromagnetic Waves Class 11 Physics: Units and Measurements

Q5 jee_main_2025_07_april_morning Displacement Current

If

in_0

denotes the permittivity of free space and

Phi_mathrmE

is the flux of the electric field through the area bounded by the closed surface, then dimension of

left(in_0fracmathrmdphi_mathrmEmathrmdtright)

are that of:

A. Electric field
B. Electric potential
C. Electric charge
D. Electric current

Solution

### Related Formula According to the Maxwell-Ampere law, displacement current

I_d

is defined as:

I_d = epsilon_0 fracmathrmdPhi_Emathrmdt

### Core Logic Since

I_d

represents a physical current, its dimensions must match those of standard conduction electric current (

[A]

or

[I]

). ### Step 1: Dimensional Analysis Let us verify using fundamental dimensions: - Permittivity of free space

[epsilon_0] = [M^-1 L^-3 T^4 A^2]

- Electric flux

[Phi_E] = [M L^3 T^-3 A^-1]

- Time

[t] = [T]

left[ epsilon_0 fracmathrmdPhi_Emathrmdt right] = [M^-1 L^-3 T^4 A^2] times frac[M L^3 T^-3 A^-1][T] = [A]

This confirms the quantity is dimensionally equivalent to Electric current. ### Pattern Recognition Sees:

epsilon_0 fracmathrmdPhi_Emathrmdt

. Shortcut: Instantly identify this as Maxwell's expression for displacement current. Displacement current has the exact same dimensions as regular conduction current. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electromagnetic Waves

Q8 jee_main_2025_29_jan_evening Propagation of Electromagnetic Waves

A plane electromagnetic wave propagates along the

+x

direction in free space. The components of the electric field,

vecE

and magnetic field,

vecB

vectors associated with the wave in Cartesian frame are:

A. $E_y, B_x$ $E_y, B_x$
B. $E_y, B_z$ $E_y, B_z$
C. $E_x, B_y$ $E_x, B_y$
D. $E_z, B_y$ $E_z, B_y$

Solution

### Related Formula

hatc = hatE times hatB

where,

hatc

= unit vector in the direction of wave propagation

hatE

= unit vector of the electric field

hatB

= unit vector of the magnetic field ### Core Logic Given that the wave propagates along the

+x

direction:

hatc = hati

Let us test the combination

E_y

and

B_z

, which implies

hatE = hatj

and

hatB = hatk

:

hatE times hatB = hatj times hatk = hati

Propagation of Electromagnetic Waves diagram for Q8 - JEE Main 2025 Evening

This cross product yields exactly the direction of propagation

+x

. Therefore,

E_y

and

B_z

are appropriate components representing the cross-orthogonal vectors of the EM wave. ### Pattern Recognition Always remember the cyclic relation

hati rightarrow hatj rightarrow hatk

. Since propagation is

hati

, our cross product must be

hatj times hatk

. Thus electric field along

y

matches magnetic field along

z

. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electromagnetic Waves

Q22 jee_main_2025_29_jan_evening Displacement Current

A parallel plate capacitor consisting of two circular plates of radius

10mathrm~cm

is being charged by a constant current of

0.15mathrm~A

. If the rate of change of potential difference between the plates is

7 times 10^8mathrm~V/s

then the integer value of the distance between the parallel plates is ______

mumathrmm

.

left(textTake epsilon_0 = 9 times 10^-12fracmathrmFmathrmm, pi = frac227right)

$$$$

Numerical Answer. Answer: 1320 to 1320

Solution

### Related Formula

I_d = I_c = C fracdVdt

C = fracepsilon_0 Ad = fracepsilon_0 pi r^2d

### Core Logic Combining the current expression with capacitance relations yields:

I = left(fracepsilon_0 pi r^2dright) fracdVdt

Isolating plate separation distance

d

:

d = fracepsilon_0 pi r^2I cdot left(fracdVdtright)

Substitute the parameters specified by the problem layout: -

r = 10mathrm~cm = 0.1mathrm~m

-

epsilon_0 = 9 times 10^-12

-

pi = 22/7

-

I = 0.15mathrm~A

-

fracdVdt = 7 times 10^8mathrm~V/s

d = frac(9 times 10^-12) times left(frac227right) times (0.1)^20.15 times (7 times 10^8)

d = frac9 times 10^-12 times 22 times 0.01 times 10^80.15 = frac198 times 10^-40.15

d = 1320 times 10^-6mathrm~m = 1320mumathrmm

Thus, the integer value for the distance is 1320. ### Pattern Recognition The charging current matches displacement current identically. Treat the system as a standard linear differential capacitor setup using

I = C fracdVdt

. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electromagnetic Waves Class 12 Physics: Electrostatic Potential and Capacitance

Solution & Explanation

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