A plane electromagnetic wave of frequency 40 MHz travels in free space in the x -direction. At some point and at some instant, the electric field E → has its maximum value of 750 N C - 1 in the y -direction. The magnitude and direction of the magnetic field will be -

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PhysicsEM WavesElectromagnetic wavesEasy2 minPYQ_2020

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A plane electromagnetic wave of frequency $40 MHz$ travels in free space in the $x$ -direction. At some point and at some instant, the electric field $\vec{E}$ has its maximum value of $750 N C^{- 1}$ in the $y$ -direction. The magnitude and direction of the magnetic field will be -

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Answer:

Solution:

Crossed electric and magnetic field are constituent of electromagnetic wave

$B_{m} = \frac{E_{m}}{C} = \frac{750}{3 \times 10^{8}} = 2.5 μT$ in $z$ -direction

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:EM WavesSubtopic:Electromagnetic waves

⏱ 2mℹ️ Source: PYQ_2020

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