Skip to content
PhysicsNEET 2025Electromagnetic spectrum

Electromagnetic Waves

The electric field in a plane electromagnetic wave is given by Ez=60cos⁡(5x+1.5×109t)E_z = 60\cos(5x + 1.5 \times 10^{9}t) V/m. Then the expression for the corresponding magnetic field is (here the subscripts denote the direction of the field):

  1. A.By=2×10−7cos⁡(5x+1.5×109t)B_y = 2 \times 10^{-7}\cos(5x + 1.5 \times 10^{9}t) T✓
  2. B.Bx=2×10−7cos⁡(5x+1.5×109t)B_x = 2 \times 10^{-7}\cos(5x + 1.5 \times 10^{9}t) T
  3. C.Bz=60cos⁡(5x+1.5×109t)B_z = 60\cos(5x + 1.5 \times 10^{9}t) T
  4. D.By=60sin⁡(5x+1.5×109t)B_y = 60\sin(5x + 1.5 \times 10^{9}t) T

Solution

B0=E0c=603×108=2×10−7B_0 = \frac{E_0}{c} = \frac{60}{3\times10^{8}} = 2\times10^{-7} T. The wave travels along −x-x (phase 5x+ωt5x + \omega t); with E⃗\vec{E} along zz, E⃗×B⃗\vec{E}\times\vec{B} must point along −x-x, which requires B⃗\vec{B} along +y+y. So By=2×10−7cos⁡(5x+1.5×109t)B_y = 2\times10^{-7}\cos(5x + 1.5\times10^{9}t) T.

Save for spaced revision (free account)

More from this chapter in past papers

From Physics › Electromagnetic Waves › Electromagnetic spectrum — lectures, notes and every past question on it.

Last revised . Found a mistake? Tell us — corrections are the fastest thing we act on.

ShareWhatsAppTelegram