A plane electromagnetic wave of frequency 25 GHz is propagating in vacuum along the z-direction. At a particular point in space and time, the magnetic field is given by B⃗ = 5 × 10⁻⁸ĵT. The corresponding electric field E⃗ is (speed of light c = 3 × 10⁸ ms –1 )

Solution

Physics sir · 3 steps

Get the size from E = cB, then fix the direction with E⃗ × B⃗ along the direction of travel.

1. Find the magnitude. In vacuum the field amplitudes are linked by E = cB. The frequency of 25 GHz is not needed for this.

   E = cB = (3 × 10⁸)(5 × 10⁻⁸) = 15 V m⁻¹

2. Fix the direction. An EM wave travels along E⃗ × B⃗. Here B⃗ is along ĵ and the wave moves along +k̂. Since î × ĵ = k̂, E⃗ must point along +î.
3. So. E⃗ = 15 î V m⁻¹, which is option (a).

The idea to keep. For an EM wave, E = cB gives the size and E⃗ × B⃗ points along the direction of travel. Use that cross product to settle the sign every time.

Answer from the NTA final answer key. Solution by PrepWiser.