A beam of light travelling along x-axis is described by the electric field E y =(600 Vm -1) sin ω( t - x / c ) then maximum magnetic force on a charge. q =2 e , moving along y-axis with a speed of 3.0 × 107 ms -1 is (e=1.6 × 10-19 C )

Question

A beam of light travelling along x-axis is described by the electric field E y =(600 Vm -1) sin ω( t - x / c ) then maximum magnetic force on a charge. q =2 e , moving along y-axis with a speed of 3.0 × 107 ms -1 is (e=1.6 × 10-19 C ) (A)  19.2× 10-17 N  (B)  1.92× 10-17 N  (C) 0.192 N (D) none of these

Tardigrade Admin · Accepted Answer

Maximum magnetic field is given by B0=(E0/c) Here, E0=600 Vm -1 c=3 × 108 m / s ∴ B0=(600/3 × 108) =2 × 10-6 T Maximum magnetic force imposed on given charge is Fm=q v B0=2 e v B0 =2 × 1.6 × 10-19 × 3 × 107 × 2 × 10-6 =1.92 × 10-17 N

Q. A beam of light travelling along $x$ -axis is described by the electric field $E_{y} = (600 V m^{- 1}) sin ω (t - x / c)$ then maximum magnetic force on a charge. $q = 2^{e}$ , moving along $y$ -axis with a speed of $3.0 \times 1 0^{7} m s^{- 1}$ is $(e = 1.6 \times 1 0^{- 19} C)$

$19.2 \times 1 0^{- 17} N$

$1.92 \times 1 0^{- 17} N$

$0.192 N$

none of these

Q. A beam of light travelling along x-axis is described by the electric field Ey​=(600Vm−1)sinω(t−x/c) then maximum magnetic force on a charge. q=2e, moving along y-axis with a speed of 3.0×107ms−1 is (e=1.6×10−19C)

19.2×10−17N

1.92×10−17N

0.192N

none of these

Maximum magnetic field is given by B0​=cE0​​ Here, E0​=600Vm−1 c=3×108m/s ∴B0​=3×108600​ =2×10−6T Maximum magnetic force imposed on given charge is Fm​=qvB0​=2evB0​ =2×1.6×10−19×3×107×2×10−6 =1.92×10−17N

Q. A beam of light travelling along $x$ -axis is described by the electric field $E_{y} = (600 V m^{- 1}) sin ω (t - x / c)$ then maximum magnetic force on a charge. $q = 2^{e}$ , moving along $y$ -axis with a speed of $3.0 \times 1 0^{7} m s^{- 1}$ is $(e = 1.6 \times 1 0^{- 19} C)$

$19.2 \times 1 0^{- 17} N$

$1.92 \times 1 0^{- 17} N$

$0.192 N$