The electric field a plane electromagnetic wave is given by : The expression for magnetic field associated with this electromagnetic wave is ____ T.

The phase of the wave is (0.6 x 103 x - 1.8 x 1011 t). This indicates the wave propagates in the +x direction, so c = i. The electric field oscillates along the y-axis, so E = j. From B = c x E, we have B = i x j = k. So, the magnetic field is along the z-axis (Bz). Wave speed v = c = (omega/k) = 1.8 x 10110.6 x 103 = 3 x 108 m/s. The amplitude of the magnetic field is: The phase remains exactly the same as the electric field: B0 = E0/c gives the magnitude. The vector identity B = v x E gives the direction. Phase part never changes sign or terms between E and B equations. Class 12 Physics: Electromagnetic Waves

The electric field a plane electromagnetic wave is given by : The expression for magnetic field associated with this electromagnetic wave is ____ T.

The phase of the wave is (0.6 x 103 x - 1.8 x 1011 t). This indicates the wave propagates in the +x direction, so c = i. The electric field oscillates along the y-axis, so E = j. From B = c x E, we have B = i x j = k. So, the magnetic field is along the z-axis (Bz). Wave speed v = c = (omega/k) = 1.8 x 10110.6 x 103 = 3 x 108 m/s. The amplitude of the magnetic field is: The phase remains exactly the same as the electric field: B0 = E0/c gives the magnitude. The vector identity B = v x E gives th

The electric field a plane electromagnetic wave is given by : The expression for magnetic field associated with this electromagnetic wave is ____ T.

The phase of the wave is (0.6 x 103 x - 1.8 x 1011 t). This indicates the wave propagates in the +x direction, so c = i. The electric field oscillates along the y-axis, so E = j. From B = c x E, we have B = i x j = k. So, the magnetic field is along the z-axis (Bz). Wave speed v = c = (omega/k) = 1.8 x 10110.6 x 103 = 3 x 108 m/s. The amplitude of the magnetic field is: The phase remains exactly the same as the electric field: B0 = E0/c gives the magnitude. The vector identity B = v x E gives the direction. Phase part never changes sign or terms between E and B equations. Class 12 Physics: Electromagnetic Waves

The electric field a plane electromagnetic wave is given by : The expression for magnetic field associated with this electromagnetic wave is ____ T.

The phase of the wave is (0.6 x 103 x - 1.8 x 1011 t). This indicates the wave propagates in the +x direction, so c = i. The electric field oscillates along the y-axis, so E = j. From B = c x E, we have B = i x j = k. So, the magnetic field is along the z-axis (Bz). Wave speed v = c = (omega/k) = 1.8 x 10110.6 x 103 = 3 x 108 m/s. The amplitude of the magnetic field is: The phase remains exactly the same as the electric field: B0 = E0/c gives the magnitude. The vector identity B = v x E gives th

Solution & Explanation

### 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

More Electromagnetic Waves Previous-Year Questions

Q18 jee_main_2025_02_april_evening Speed of Electromagnetic Waves

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

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]

[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

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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.

Solution & Explanation

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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.$$E_y = 69 \sin [ 0.6 \times 10^3 x - 1.8 \times 10^{11} t ]\text{ V/m}.$$ The expression for magnetic field associated with this electromagnetic wave is ____ T.

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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.