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Electromagnetic Waves

17 JEE Main questions · Physics

  1. PhysicsJEE Main 21 Jan 2026, Shift 1·Q10·MCQ·Easy–moderate
    Open

    EM waves

    The electric field of a plane electromagnetic wave is given by: Ey=69sin⁡[0.6×103x−1.8×1011t] V/m.E_y = 69\sin[0.6\times10^{3}x - 1.8\times10^{11}t]\ \mathrm{V/m}. The expression for magnetic field associated with this electromagnetic wave is T.
    1. (1)Bz=2.3×10−7sin⁡[0.6×103x−1.8×1011t]B_z = 2.3\times10^{-7}\sin[0.6\times10^{3}x - 1.8\times10^{11}t]
    2. (2)Bz=2.3×10−7sin⁡[0.6×103x+1.8×1011t]B_z = 2.3\times10^{-7}\sin[0.6\times10^{3}x + 1.8\times10^{11}t]
    3. (3)By=69sin⁡[0.6×103x+1.8×1011t]B_y = 69\sin[0.6\times10^{3}x + 1.8\times10^{11}t]
    4. (4)By=2.3×10−7sin⁡[0.6×103x−1.8×1011t]B_y = 2.3\times10^{-7}\sin[0.6\times10^{3}x - 1.8\times10^{11}t]
    We corrected a printing error in the original paper

    Source printed “The electric field a plane electromagnetic wave”.

  2. PhysicsJEE Main 21 Jan 2026, Shift 2·Q21·Numerical·Easy–moderate
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    EM waves

    An electromagnetic wave of frequency 100 MHz propagates through a medium of conductivity, σ=10\sigma = 10 mho/m. The ratio of maximum conducting current density to maximum displacement current density is . [Take 14πϵ0=9×109 Nm2/C2]\left[\text{Take } \dfrac{1}{4\pi\epsilon_0} = 9\times10^{9}\ \mathrm{Nm^2/C^2}\right]
  3. PhysicsJEE Main 22 Jan 2026, Shift 1·Q22·Numerical·Easy–moderate
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    EM waves

    The electric field of a plane electromagnetic wave, travelling in an unknown non-magnetic medium is given by, Ey=20sin⁡(3×106x−4.5×1014t) V/mE_y = 20\sin\left(3\times10^{6}x - 4.5\times10^{14}t\right)\ \mathrm{V/m} (where x, t and other values have S.I. units). The dielectric constant of the medium is . (speed of light in free space is 3×1083\times10^{8} m/s)
  4. PhysicsJEE Main 22 Jan 2026, Shift 2·Q18·MCQ·Easy–moderate
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    EM waves

    A laser beam has intensity of 4.0×1014 W/m24.0\times10^{14}\ \mathrm{W/m^2}. The amplitude of magnetic field associated with beam is T. (Take ϵ0=8.85×10−12 C2/Nm2\epsilon_0 = 8.85\times10^{-12}\ \mathrm{C^2/Nm^2} and c=3×108c = 3\times10^{8} m/s)
    1. (1)2.0
    2. (2)18.3
    3. (3)5.5
    4. (4)1.83
  5. PhysicsJEE Main 23 Jan 2026, Shift 1·Q3·MCQ·Easy
    Open

    EM waves

    Match List-I with List-II.
    List-I (Relation)List-II (Law)
    A.∮E⃗⋅dl⃗=−ddt∮B⃗⋅da⃗\displaystyle\oint\vec E\cdot d\vec l = -\frac{d}{dt}\oint\vec B\cdot d\vec aI.Ampere's circuital law.
    B.∮B⃗⋅dl⃗=μ0(I+ϵ0dϕEdt)\displaystyle\oint\vec B\cdot d\vec l = \mu_0\left(I + \epsilon_0\frac{d\phi_E}{dt}\right)II.Faraday's laws of electromagnetic induction.
    (C)∮E⃗⋅da⃗=1ϵ0∫ρ dv\displaystyle\oint\vec E\cdot d\vec a = \frac{1}{\epsilon_0}\int\rho\,dvIII.Ampere-Maxwell law
    (D)∮B⃗⋅dl⃗=μ0I\displaystyle\oint\vec B\cdot d\vec l = \mu_0 IIV.Gauss's law of electrostatics
    Choose the correct answer from the options given below:
    1. (1)A-II, B-III, C-I, D-IV
    2. (2)A-II, B-III, C-IV, D-I
    3. (3)A-I, B-IV, C-III, D-II
    4. (4)A-IV, B-I, C-II, D-III
    We corrected a printing error in the original paper

    Source printed row B as “∮B⋅dl⃗=μ0(1+ϵ0dϕEdt)\oint B\cdot d\vec l = \mu_0\left(1 + \epsilon_0\frac{d\phi_E}{dt}\right)” (digit 1 for the current II, no arrow on BB).

  6. PhysicsJEE Main 23 Jan 2026, Shift 1·Q22·Numerical·Easy–moderate
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    EM waves

    The equation of the electric field of an electromagnetic wave propagating through free space is given by: E=377 sin⁡(6.27×103t−2.09×10−5x) N/CE = \sqrt{377}\,\sin(6.27\times10^{3}t - 2.09\times10^{-5}x)\ \mathrm{N/C} The average power of the electromagnetic wave is (1α)W/m2\left(\dfrac{1}{\alpha}\right)\mathrm{W/m^2}. The value of α\alpha is

    (Take μ0ε0=377 in SI units)\left(\text{Take } \sqrt{\dfrac{\mu_0}{\varepsilon_0}} = 377 \text{ in SI units}\right)

    We corrected a printing error in the original paper

    Source printed “he average power”.

  7. PhysicsJEE Main 23 Jan 2026, Shift 2·Q18·MCQ·Easy–moderate
    Open

    EM waves

    The ratio of speeds of electromagnetic waves in vacuum and a medium, having dielectric constant k=3k = 3 and permeability of μ=2μ0\mu = 2\mu_0, is
    (μ0\mu_0 = permeability of vacuum)
    1. (1)36:136:1
    2. (2)3:23:2
    3. (3)6:16:1
    4. (4)6:1\sqrt6:1
  8. PhysicsJEE Main 24 Jan 2026, Shift 1·Q10·MCQ·Easy
    Open

    EM waves

    Match the List-I with List-II
    List-IList-II
    ARadio-waveIis produced by Magnetron valve
    BMicro-waveIIDue to change in the vibrational modes of atoms
    CInfrared-waveIIIDue to inner shell electrons moving from higher energy level to lower energy level
    DX-rayIVDue to rapid acceleration of electrons
    Choose the correct answer from the options given below:
    1. (1)(A)-(II), (B)-(IV), (C)-(III), (D)-(I)
    2. (2)(A)-(IV), (B)-(III), (C)-(I), (D)-(II)
    3. (3)(A)-(IV), (B)-(I), (C)-(II), (D)-(III)
    4. (4)(A)-(IV), (B)-(II), (C)-(I), (D)-(III)
    We corrected a printing error in the original paper

    Source printed “Magnetron value”.

  9. PhysicsJEE Main 28 Jan 2026, Shift 1·Q7·MCQ·Easy–moderate
    Open

    EM waves

    The electric field of an electromagnetic wave travelling through a medium is given by E⃗(x,t)=25sin⁡(2.0×1015t−107x) n^\vec E(x,t) = 25\sin(2.0\times10^{15}t - 10^{7}x)\,\hat n then the refractive index of the medium is .
    (All given measurements are in SI units)
    1. (1)1.2
    2. (2)2
    3. (3)1.5
    4. (4)1.7
    We corrected a printing error in the original paper

    Source printed “All given measurement are”.

  10. PhysicsJEE Main 28 Jan 2026, Shift 2·Q2·MCQ·Easy–moderate
    Open

    EM waves

    A plane electromagnetic wave is moving in free space with velocity c=3×108c = 3\times10^{8} m/s and its electric field is given as E⃗=54sin⁡(kz−ωt) j^ V/m\vec E = 54\sin(kz - \omega t)\,\hat j\ \mathrm{V/m}, where j^\hat j is the unit vector along y-axis. The magnetic field vector B⃗\vec B of the wave is:
    1. (1)−1.8×10−7sin⁡(kz−ωt) i^ T-1.8\times10^{-7}\sin(kz - \omega t)\,\hat i\ \mathrm{T}
    2. (2)1.4×10−7sin⁡(kz−ωt) k^ T1.4\times10^{-7}\sin(kz - \omega t)\,\hat k\ \mathrm{T}
    3. (3)1.4×10−7sin⁡(kz−ωt) i^ T1.4\times10^{-7}\sin(kz - \omega t)\,\hat i\ \mathrm{T}
    4. (4)+1.8×10−7sin⁡(kz−ωt) i^ T+1.8\times10^{-7}\sin(kz - \omega t)\,\hat i\ \mathrm{T}
  11. PhysicsJEE Main 2 Apr 2026, Shift 1·Q18·MCQ·Easy–moderate
    Open

    EM waves

    An electromagnetic wave travelling in x-direction is described by field equation Ey=300sin⁡ω(t−xc).E_y = 300\sin\omega\left(t - \frac{x}{c}\right). If the electron is restricted to move in y-direction only with speed of 1.5×1061.5\times10^{6} m/s then ratio of maximum electric and magnetic forces acting on the electron is .
    1. (1)200
    2. (2)150
    3. (3)400
    4. (4)300
  12. PhysicsJEE Main 2 Apr 2026, Shift 2·Q12·MCQ·Easy–moderate
    Open

    EM waves

    An electromagnetic wave travels in free space along the xx-direction. At a particular point in space and time, B⃗=2×10−7 ȷ^\vec{B} = 2\times10^{-7}\,\hat{\jmath} T is associated with this wave. The value of corresponding electric field E⃗\vec{E} at this point is V/m.
    1. (1)60 k^60\,\hat{k}
    2. (2)−60 k^-60\,\hat{k}
    3. (3)30 k^30\,\hat{k}
    4. (4)−600 k^-600\,\hat{k}
  13. PhysicsJEE Main 4 Apr 2026, Shift 2·Q15·MCQ·Easy–moderate
    Open

    EM waves

    A magnetic field vector in an electromagnetic wave is represented by, B⃗=B0sin⁡(2πνt−2πxλ)j^.\vec B = B_0\sin\left(2\pi\nu t - \frac{2\pi x}{\lambda}\right)\hat j. Its associated electric field vector is .
    1. (1)E⃗=−νλB0sin⁡(2πνt−2πxλ)k^\vec E = -\nu\lambda B_0\sin\left(2\pi\nu t - \dfrac{2\pi x}{\lambda}\right)\hat k
    2. (2)E⃗=−νλB0sin⁡(2πνt−2πxλ)i^\vec E = -\nu\lambda B_0\sin\left(2\pi\nu t - \dfrac{2\pi x}{\lambda}\right)\hat i
    3. (3)E⃗=νλB0sin⁡(2πνt−2πxλ)k^\vec E = \nu\lambda B_0\sin\left(2\pi\nu t - \dfrac{2\pi x}{\lambda}\right)\hat k
    4. (4)E⃗=νλB0sin⁡(2πνt−2πxλ)i^\vec E = \nu\lambda B_0\sin\left(2\pi\nu t - \dfrac{2\pi x}{\lambda}\right)\hat i
    We corrected a printing error in the original paper

    Source printed the frequency as a Latin “v” (2πvt2\pi\mathrm{v}t, vλB0\mathrm{v}\lambda B_0) instead of ν\nu.

  14. PhysicsJEE Main 5 Apr 2026, Shift 1·Q12·MCQ·Easy–moderate
    Open

    EM waves

    A displacement current of 4.0 A can be set up in the space between two parallel plates of 6 μF6\ \mu\mathrm{F} capacitor. The rate of change of potential difference across the plates of the capacitor is nearly α×106\alpha\times10^{6} V/s. The value of α\alpha is .
    1. (1)0.58
    2. (2)0.67
    3. (3)0.82
    4. (4)0.75
  15. PhysicsJEE Main 5 Apr 2026, Shift 2·Q14·MCQ·Easy
    Open

    EM waves

    Given below are two statements : one is labelled as Assertion (A) : and the other is labelled as Reason (R).

    Assertion (A) : The electromagnetic wave exerts pressure on the surface on which it is allowed to fall.

    Reason (R) : There is no mass associated with the electromagnetic waves.

    In the light of the above statements, choose the correct answer from the options given below :

    1. (1)Both (A) and (R) are true and (R) is the correct explanation of (A)
    2. (2)Both (A) and (R) are true but (R) is not the correct explanation of (A)
    3. (3)(A) is true but (R) is false
    4. (4)(A) is false but (R) is true
    We corrected a printing error in the original paper

    Source printed “on which they are allowed to fall”.

  16. PhysicsJEE Main 6 Apr 2026, Shift 1·Q15·MCQ·Easy
    Open

    EM waves

    A point light source emits E.M. waves in free space. A detector, placed at a distance of L m, measures the intensity as I0I_0. The detector is now shifted to another location on the same spherical surface ensuring the angle between original location and new location as 45∘45^\circ. The measured intensity at new location will be .
    1. (1)I04\dfrac{I_0}{4}
    2. (2)I0I_0
    3. (3)I02\dfrac{I_0}{\sqrt2}
    4. (4)I02\dfrac{I_0}{2}
  17. PhysicsJEE Main 6 Apr 2026, Shift 2·Q6·MCQ·Easy
    Open

    EM waves

    For an electromagnetic wave propagating through vacuum, k⃗\vec k, E⃗\vec E and ω\omega represent propagation vector, electric field and angular frequency, respectively. The magnetic field associated with this wave is represented by:
    1. (1)E⃗×k⃗ω\dfrac{\vec E\times\vec k}{\omega}
    2. (2)k⃗×E⃗ω\dfrac{\vec k\times\vec E}{\omega}
    3. (3)ω(E⃗×k⃗)\omega\left(\vec E\times\vec k\right)
    4. (4)ω(k⃗×E⃗)\omega\left(\vec k\times\vec E\right)