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MCQEasyJEE 2024Characteristics of EM Waves

JEE Physics 2024 Question with Solution

The magnetic field in a plane electromagnetic wave is:

By=(3.5×10−7)sin⁡(1.5×103x+0.5×1011t) TB_y = (3.5 \times 10^{-7}) \sin(1.5 \times 10^3 x + 0.5 \times 10^{11} t) \, \text{T}. The corresponding electric field will be:

  • A

    Ey=1.17sin⁡(1.5×103x+0.5×1011t) V/mE_y = 1.17 \sin(1.5 \times 10^3 x + 0.5 \times 10^{11} t) \, \text{V/m}

  • B

    Ez=105sin⁡(1.5×103x+0.5×1011t) V/mE_z = 105 \sin(1.5 \times 10^3 x + 0.5 \times 10^{11} t) \, \text{V/m}

  • C

    Ez=1.17sin⁡(1.5×103x+0.5×1011t) V/mE_z = 1.17 \sin(1.5 \times 10^3 x + 0.5 \times 10^{11} t) \, \text{V/m}

  • D

    Ey=10.5sin⁡(1.5×103x+0.5×1011t) V/mE_y = 10.5 \sin(1.5 \times 10^3 x + 0.5 \times 10^{11} t) \, \text{V/m}

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given:

  • By=(3.5×10−7)sin⁡(1.5×103x+0.5×1011t) TB_y = (3.5 \times 10^{-7}) \sin(1.5 \times 10^3 x + 0.5 \times 10^{11} t) \, \text{T}
  • Speed of light, c=3×108 m/sc = 3 \times 10^8 \, \text{m/s}

Find: The corresponding electric field.

For a plane electromagnetic wave, the magnitudes of electric and magnetic fields are related by

E=cBE = cB

Substituting the given magnetic field amplitude,

E=(3×108)(3.5×10−7)E = (3 \times 10^8)(3.5 \times 10^{-7}) E=1.05×102=105E = 1.05 \times 10^2 = 105

So the magnitude of the electric field is

E=105sin⁡(1.5×103x+0.5×1011t) V/mE = 105 \sin(1.5 \times 10^3 x + 0.5 \times 10^{11} t) \, \text{V/m}

Since the magnetic field is along the yy-direction, the electric field must be perpendicular to both the direction of propagation and the magnetic field. Therefore, the electric field is along the zz-direction.

Therefore, the corresponding electric field is Ez=105sin⁡(1.5×103x+0.5×1011t) V/mE_z = 105 \sin(1.5 \times 10^3 x + 0.5 \times 10^{11} t) \, \text{V/m}. The correct option is B.

Direct Amplitude Relation

Given: B0=3.5×10−7 TB_0 = 3.5 \times 10^{-7} \, \text{T}

Find: The amplitude and direction of the electric field.

Use the direct electromagnetic wave relation

E0=cB0E_0 = cB_0

So,

E0=(3×108)(3.5×10−7)=105 V/mE_0 = (3 \times 10^8)(3.5 \times 10^{-7}) = 105 \, \text{V/m}

The field is not along yy because electric and magnetic fields are mutually perpendicular in an electromagnetic wave. Hence the electric field must be along zz.

Therefore, the correct field is Ez=105sin⁡(1.5×103x+0.5×1011t) V/mE_z = 105 \sin(1.5 \times 10^3 x + 0.5 \times 10^{11} t) \, \text{V/m}.

Common mistakes

  • Using E=BcE = \frac{B}{c} instead of E=cBE = cB. This gives a much smaller value and is dimensionally incorrect for electromagnetic waves. Always use the relation E/B=cE/B = c.

  • Choosing the electric field along the yy-direction. In a plane electromagnetic wave, E⃗\vec{E} and B⃗\vec{B} are perpendicular to each other, so if B⃗\vec{B} is along yy, E⃗\vec{E} cannot also be along yy.

  • Calculating the amplitude correctly but forgetting that the sinusoidal phase remains the same. The electric and magnetic fields of the wave have the same phase, so the argument of the sine function must remain unchanged.

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