MCQEasyJEE 2023Characteristics of EM Waves

JEE Physics 2023 Question with Solution

If E\vec{E} represents the electric field vector and K\vec{K} the propagation vector of electromagnetic waves in vacuum, then the magnetic field vector is given by (ω\omega - angular frequency):

  • A

    B=K×Eω\vec{B} = \frac{\vec{K} \times \vec{E}}{\omega}

  • B

    B=ω(E×K)\vec{B} = \omega (\vec{E} \times \vec{K})

  • C

    B=ω(K×E)\vec{B} = \omega (\vec{K} \times \vec{E})

  • D

    (K×E)(\vec{K} \times \vec{E})

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: E\vec{E} is the electric field vector, K\vec{K} is the propagation vector, and ω\omega is the angular frequency of an electromagnetic wave in vacuum.

Find: The correct expression for the magnetic field vector B\vec{B}.

For an electromagnetic wave in vacuum, the electric field, magnetic field, and propagation direction are mutually perpendicular. The magnetic field direction is obtained from the cross product of the propagation vector and the electric field.

B=K×Eω\vec{B} = \frac{\vec{K} \times \vec{E}}{\omega}

This also gives the correct vector direction because K×E\vec{K} \times \vec{E} is perpendicular to both K\vec{K} and E\vec{E}.

  • Option A is correct.
  • Option B has the cross product in the reverse order, so the direction is wrong.
  • Option C is dimensionally incorrect because it multiplies by ω\omega instead of dividing by it.
  • Option D is incomplete because it does not include division by ω\omega.

Therefore, the correct option is A.

Common mistakes

  • Reversing the cross product as E×K\vec{E} \times \vec{K}. This is wrong because cross products are order-sensitive and reversing the order changes the direction. Use K×E\vec{K} \times \vec{E} to get the correct magnetic field direction.

  • Multiplying by ω\omega instead of dividing by it. This is wrong because it makes the expression dimensionally inconsistent. Check the standard EM wave relation carefully before choosing the option.

  • Assuming any vector perpendicular to E\vec{E} is acceptable. This is wrong because B\vec{B} must be perpendicular to both E\vec{E} and K\vec{K}. Use the cross product to satisfy both direction conditions together.

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