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Question: The magnetic field of a beam emerging from a filter facing a flood light as given by \(B = 12 \time...

The magnetic field of a beam emerging from a filter facing a flood light as given by

B=12×10−8sin⁡(1.02×107z−3.60×1015t)T.B = 12 \times 10^{- 8}\sin(1.02 \times 10^{7}z - 3.60 \times 10^{15}t)T. The average intensity of the beam is:

A

1.71Wm−21.71Wm^{- 2}

B

2.1Wm−22.1Wm^{- 2}

C

3.2Wm−23.2Wm^{- 2}

D

2.9Wm−22.9Wm^{- 2}

Answer

1.71Wm−21.71Wm^{- 2}

Explanation

Solution

: Here, B=12×10−8B = 12 \times 10^{- 8}sin (1.20×107z−3.6×1015t)(1.20 \times 10^{7}z - 3.6 \times 10^{15}t) T comparing it with, B=B0sin⁡(kz−ωt),B = B_{0}\sin(kz - \omega t),

we have B0=12×10−8\mathbf{B}_{0} = 12 \times 10^{- 8}T

∴Iav=12B02cμ0\therefore I_{av} = \frac{1}{2}\frac{B_{0}^{2}c}{\mu_{0}}

=12×(12×10−8)2×3×1084π×10−7= \frac{1}{2} \times \frac{(12 \times 10^{- 8})^{2} \times 3 \times 10^{8}}{4\pi \times 10^{- 7}}

=1.71Wm−2= 1.71Wm^{- 2}