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Physics Question 37 – JEE-MAIN 2026

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

In an electromagnetic wave, the electric field E→, magnetic field B→, and the direction of propagation k→ are mutually perpendicular.

Step 1: Calculate the magnitude of the electric field✦ Active

The magnitude of the electric field E is related to the magnitude of the magnetic field B by the speed of light c in free space (c=3×108 m/s). Given B=2×10−7 T.

E=cB=(3×108 m/s)×(2×10−7 T)=60 V/m
Step 2: Determine the direction of the electric field○ Expand

The electromagnetic wave propagates along the x-direction (i^). The magnetic field is in the y-direction (j^). The direction of propagation is given by k^propagation=E^×B^. So, i^=E^×j^. Using the right-hand rule for cross products, if E^ is in the −k^ direction, then (−k^)×j^=−(k^×j^)=−(−i^)=i^. Thus, the electric field is in the −k^ direction.

💡 Teacher's Secret Hint

Remember the cyclic order of unit vectors: i^×j^=k^, j^×k^=i^, k^×i^=j^.

Step 3: Combine magnitude and direction to find E→○ Expand

Combining the magnitude E=60 V/m and the direction −k^, the electric field vector is:

E→=60(−k^)=−60k^ V/m
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