If mu_0 and varepsilon_0 are the permeability and permittivity of free space, respectively, then the dimension of left(frac1mu_0varepsilon_0right) is:

Solution & Explanation

### 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

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Electromagnetic Waves

More Electromagnetic Waves Previous-Year Questions

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
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
  • B. E_y, B_z
  • C. E_x, B_y
  • D. 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
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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