Electromagnetic Waves

JEE Main level, single correct, medium

8 of 165

Question 8 of 165

An electromagnetic wave travels in free space along the xx-direction. At a particular point in space and time, B⃗=2×10−7j^\vec{B} = 2 \times 10^{-7} \hat{j} T is associated with this wave. The value of corresponding electric field E⃗\vec{E} at this point is _______ V/m.

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Answer: (B) −60k^-60 \hat{k}

The direction of propagation of an electromagnetic wave is given by the direction of E⃗×B⃗\vec{E} \times \vec{B}.

Given that the wave travels along the xx-direction, the direction of propagation is i^\hat{i}.

The magnetic field is given as B⃗=2×10−7j^\vec{B} = 2 \times 10^{-7} \hat{j} T, so its direction is j^\hat{j}.

Let the direction of the electric field be n^\hat{n}. We must have:

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

Since (−k^)×j^=i^(-\hat{k}) \times \hat{j} = \hat{i}, the direction of the electric field is −k^-\hat{k}.

The magnitude of the electric field is related to the magnetic field by E=cBE = cB, where c=3×108c = 3 \times 10^8 m/s is the speed of light in free space.

E=(3×108)×(2×10−7)=60E = (3 \times 10^8) \times (2 \times 10^{-7}) = 60 V/m

Therefore, the electric field vector is E⃗=−60k^\vec{E} = -60 \hat{k} V/m.

Answer: −60k^-60 \hat{k}

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