Light wave travelling in air along x-direction is given by Ey=540 sin π × 104(x-c t) Vm -1. Then, the peak value of magnetic field of wave will be (Given c =3 × 108 ms -1 )

Question

Light wave travelling in air along x-direction is given by Ey=540 sin π × 104(x-c t) Vm -1. Then, the peak value of magnetic field of wave will be (Given c =3 × 108 ms -1 ) (A) 18 × 10-7 T (B) 54 × 10-7 T (C) 54 × 10-8 T (D) 18 × 10-8 T

Tardigrade Admin · Accepted Answer

E y =540 sin π × 104( x - ct ) Vm -1 E 0=540 Vm -1 B 0=( E 0/ C )=(540/3 × 108)=18 × 10-7 T

Q. Light wave travelling in air along $x$ -direction is given by $E_{y} = 540 sin π \times 1 0^{4} (x - c t) V m^{- 1}$ . Then, the peak value of magnetic field of wave will be (Given $c = 3 \times 1 0^{8} m s^{- 1}$ )

$18 \times 1 0^{- 7} T$

$54 \times 1 0^{- 7} T$

$54 \times 1 0^{- 8} T$

$18 \times 1 0^{- 8} T$

$E_{y} = 540 sin π \times 1 0^{4} (x - c t) V m^{- 1}$
$E_{0} = 540 V m^{- 1}$
$B_{0} = \frac{E _{0}}{C} = \frac{540}{3 \times 1 0 ^{8}} = 18 \times 1 0^{- 7} T$

Q. Light wave travelling in air along x-direction is given by Ey​=540sinπ×104(x−ct)Vm−1. Then, the peak value of magnetic field of wave will be (Given c=3×108ms−1 )

18×10−7T

54×10−7T

54×10−8T

18×10−8T