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Topic: EM waves, characteristics and transverse nature· Electromagnetic Waves

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

1. \(B_y = 2\times10^{-7}\cos(5x+1.5\times10^9 t)T\)

2. \(B_x = 2\times10^{-7}\cos(5x+1.5\times10^9 t)T\)

3. \(B_z = 60\cos(5x+1.5\times10^9 t)T\)

4. \(B_y = 60\sin(5x+1.5\times10^9 t)T\)

1 answer

Best answer

Answer: 1. \(B_y = 2\times10^{-7}\cos(5x+1.5\times10^9 t)T\)

Key idea: in an EM wave, \(B_0=E_0/c\), and \(\vec E\times\vec B\) points along the direction of propagation.

The argument \(5x+1.5\times10^{9}t\) means the wave travels in the \(-x\) direction, with \(c=\omega/k=\dfrac{1.5\times10^9}{5}=3\times10^{8}\) m/s.

$$B_0=\frac{E_0}{c}=\frac{60}{3\times10^{8}}=2\times10^{-7}\ \mathrm T$$

Direction: \(E\) is along \(z\) and propagation is along \(-x\). Since \(\hat z\times\hat y=-\hat x\), the magnetic field is along \(y\). So \(B_y=2\times10^{-7}\cos(5x+1.5\times10^9t)\) T. Option 3 has the wrong magnitude and direction, and option 4 has the wrong phase.

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report
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