A beam of light travelling along X-axis is described by the electric field E y =900 sin ω( t - x / c ). The ratio of electric force to magnetic force on a charge q moving along Y-axis with a speed of 3 × 107 ms -1 will be: [Given speed of light =3 × 108 ms -1 ]

Question

A beam of light travelling along X-axis is described by the electric field E y =900 sin ω( t - x / c ). The ratio of electric force to magnetic force on a charge q moving along Y-axis with a speed of 3 × 107 ms -1 will be: [Given speed of light =3 × 108 ms -1 ] (A) 1: 1 (B) 1: 10 (C) 10: 1 (D) 1: 2

Tardigrade Admin · Accepted Answer

E y =900 sin (ω t -(ω x / c )) E 0=900 <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/98964f77a515648c5714b2f0d4f1c28b-.png" /> F E = qE 0 F B = qvB 0 ( F E / F B )=( E 0/ vB 0)=( c / v )=(3 × 108/3 × 107)=10: 1

$1 : 1$

$1 : 10$

$10 : 1$

$1 : 2$

$E_{y} = 900 sin (ω t - \frac{ω x}{c})$
$E_{0} = 900$

$F_{E} = q E_{0}$
$F_{B} = q v B_{0}$
$\frac{F _{E}}{F _{B}} = \frac{E _{0}}{v B _{0}} = \frac{c}{v} = \frac{3 \times 1 0 ^{8}}{3 \times 1 0 ^{7}} = 10 : 1$

Q. A beam of light travelling along X-axis is described by the electric field Ey​=900sinω(t−x/c). The ratio of electric force to magnetic force on a charge q moving along Y-axis with a speed of 3×107ms−1 will be : [Given speed of light =3×108ms−1 ]

1:1

1:10

10:1

1:2

Ey​=900sin(ωt−cωx​) E0​=900 FE​=qE0​ FB​=qvB0​ FB​FE​​=vB0​E0​​=vc​=3×1073×108​=10:1

$1 : 1$

$1 : 10$

$10 : 1$

$1 : 2$

$E_{y} = 900 sin (ω t - \frac{ω x}{c})$
$E_{0} = 900$

$F_{E} = q E_{0}$
$F_{B} = q v B_{0}$
$\frac{F _{E}}{F _{B}} = \frac{E _{0}}{v B _{0}} = \frac{c}{v} = \frac{3 \times 1 0 ^{8}}{3 \times 1 0 ^{7}} = 10 : 1$