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MCQEasyJEE 2025Characteristics of EM Waves

JEE Physics 2025 Question with Solution

A plane electromagnetic wave of frequency 20 MHz20 \, \text{MHz} travels in free space along the +x+x direction. At a particular point in space and time, the electric field vector of the wave is Ey=9.3 V/mE_y = 9.3 \, \text{V/m}. Then, the magnetic field vector of the wave at that point is:

  • A

    Bz=9.3×10−8 TB_z = 9.3 \times 10^{-8} \, \text{T}

  • B

    Bz=1.55×10−8 TB_z = 1.55 \times 10^{-8} \, \text{T}

  • C

    Bz=6.2×10−8 TB_z = 6.2 \times 10^{-8} \, \text{T}

  • D

    Bz=3.1×10−8 TB_z = 3.1 \times 10^{-8} \, \text{T}

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: A plane electromagnetic wave travels in free space along +x+x. At the given point, Ey=9.3 V/mE_y = 9.3 \, \text{V/m}.

Find: The magnetic field vector component BzB_z at that point.

For an electromagnetic wave in free space, the magnitudes of electric and magnetic fields satisfy

E=cBE = cB

where c=3×108 m/sc = 3 \times 10^8 \, \text{m/s}.

So,

B=EcB = \frac{E}{c}

Substituting the given value,

B=9.33×108B = \frac{9.3}{3 \times 10^8} B=3.1×10−8 TB = 3.1 \times 10^{-8} \, \text{T}

Since the wave propagates along +x+x and the electric field is along yy, the magnetic field must be along zz so that E⃗×B⃗\vec{E} \times \vec{B} points along +x+x.

Therefore, the magnetic field vector is Bz=3.1×10−8 TB_z = 3.1 \times 10^{-8} \, \text{T}. The correct option is D.

The frequency 20 MHz20 \, \text{MHz} is given, but it is not needed here because the relation between field magnitudes in free space is directly B=EcB = \frac{E}{c}.

Direction Check with Wave Propagation

Given: Wave propagation is along +x+x, and the electric field component is EyE_y.

Find: The corresponding magnetic field vector.

In a plane electromagnetic wave,

E⃗⊥B⃗⊥direction of propagation\vec{E} \perp \vec{B} \perp \text{direction of propagation}

Also,

E⃗×B⃗\vec{E} \times \vec{B}

points in the direction of propagation.

Here E⃗\vec{E} is along +y+y and propagation is along +x+x. Therefore B⃗\vec{B} must be along +z+z because

j^×k^=i^\hat{j} \times \hat{k} = \hat{i}

Now use the magnitude relation,

B=Ec=9.33×108=3.1×10−8 TB = \frac{E}{c} = \frac{9.3}{3 \times 10^8} = 3.1 \times 10^{-8} \, \text{T}

Hence, the magnetic field vector is Bz=3.1×10−8 TB_z = 3.1 \times 10^{-8} \, \text{T} and the correct option is D.

Common mistakes

  • Using the frequency 20 MHz20 \, \text{MHz} in the calculation of BB is unnecessary here. The field magnitude relation in free space is directly B=EcB = \frac{E}{c}, so use the given electric field and the speed of light.

  • Confusing E=cBE = cB with E=BcE = \frac{B}{c} leads to an incorrect order of magnitude. Rearranging correctly gives B=EcB = \frac{E}{c}, so the magnetic field is much smaller numerically than the electric field value.

  • Ignoring the direction of the magnetic field is a conceptual error. Since the wave travels along +x+x and E⃗\vec{E} is along yy, B⃗\vec{B} must be along zz so that E⃗×B⃗\vec{E} \times \vec{B} points along +x+x.

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