The magnetic field of a beam emerging from a filter facing a flood light as given by B = 12 × 10-8 sin (1.20 × 107 z - 3.60 × 1015 t) T. The average intensity of the beam is

Question

The magnetic field of a beam emerging from a filter facing a flood light as given by B = 12 × 10-8 sin (1.20 × 107 z - 3.60 × 1015 t) T. The average intensity of the beam is (A) 1.71 W m-2 (B) 2.1 W m-2 (C) 3.2 W m-2 (D) 2.9 W m-2

Tardigrade Admin · Accepted Answer

Here, B = 12 × 10-8 sin (1.20 ×107z -3.6 × 1015 t) T Comparing it with, B = B0 sin (kz - ω t), we have B0 = 12 × 10-8 T ∴ Iav = (1/2) (B20c/μ0) = (1/2) × ((12 × 10-8)2 × 3 × 108/4π × 10-7) = 1.71 W m-2

Q. The magnetic field of a beam emerging from a filter facing a flood light as given by $B = 12 \times 1 0^{- 8} s in (1.20 \times 1 0^{7} z - 3.60 \times 1 0^{15} t) T$ . The average intensity of the beam is

$1.71 W m^{- 2}$

$2.1 W m^{- 2}$

$3.2 W m^{- 2}$

$2.9 W m^{- 2}$

Q. The magnetic field of a beam emerging from a filter facing a flood light as given by B=12×10−8sin(1.20×107z−3.60×1015t)T. The average intensity of the beam is

1.71Wm−2

2.1Wm−2

3.2Wm−2

2.9Wm−2

Here, B=12×10−8sin(1.20×107z−3.6×1015t)T Comparing it with, B=B0​sin(kz−ωt), we have B0​=12×10−8T ∴Iav​=21​μ0​B02​c​ =21​×4π×10−7(12×10−8)2×3×108​ =1.71Wm−2

Q. The magnetic field of a beam emerging from a filter facing a flood light as given by $B = 12 \times 1 0^{- 8} s in (1.20 \times 1 0^{7} z - 3.60 \times 1 0^{15} t) T$ . The average intensity of the beam is

$1.71 W m^{- 2}$

$2.1 W m^{- 2}$

$3.2 W m^{- 2}$

$2.9 W m^{- 2}$