The magnetic field of a beam emerging from a filter facing a flood light as given byB=12×10-8 sin1.20×107z -3.60×1015 tT. The average intensity of the beam is:

Here, B=12×10-8 sin1.20×107 z -3.60×1015 tTComparing it with, B=B0 sinkz-ωt, we have B0=12×10-8 T∴ Iav=12B02cμ0=12×12×10-82×3×1084π×10-7=1.72 W m-2

The magnetic field of a beam emerging from a filter facing a flood light as given byB=12×10-8 sin1.20×107z -3.60×1015 tT. The average intensity of the beam is:

Here, B=12×10-8 sin1.20×107 z -3.60×1015 tTComparing it with, B=B0 sinkz-ωt, we have B0=12×10-8 T∴ Iav=12B02cμ0=12×12×10-82×3×1084π×10-7=1.72 W m-2

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The magnetic field of a beam emerging from a filter facing a flood light as given by $B = 12 \times 10^{- 8} \sin (1.20 \times 10^{7} z - 3.60 \times 10^{15} t) T$ . The average intensity of the beam is:

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$1.72 W m^{- 2}$

$2.1 W m^{- 2}$

$3.2 W m^{- 2}$

$2.9 W m^{- 2}$

answer is A.

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Detailed Solution

Here, $B = 12 \times 10^{- 8} \sin (1.20 \times 10^{7} z - 3.60 \times 10^{15} t) T$

Comparing it with, $B = B_{0} \sin (kz - ωt)$ ,

we have $B_{0} = 12 \times 10^{- 8} T$

$∴ I_{av} = \frac{1}{2} \frac{{B_{0}}^{2} c}{μ_{0}} = \frac{1}{2} \times \frac{{(12 \times 10^{- 8})}^{2} \times 3 \times 10^{8}}{4 π \times 10^{- 7}} = 1.72 W m^{- 2}$

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The magnetic field of a beam emerging from a filter facing a flood light as given byB=12×10-8 sin1.20×107z -3.60×1015 tT. The average intensity of the beam is:

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The magnetic field of a beam emerging from a filter facing a flood light as given by $B = 12 \times 10^{- 8} \sin (1.20 \times 10^{7} z - 3.60 \times 10^{15} t) T$ . The average intensity of the beam is: