Question

# A light beam travelling in the X-direction is described by the electric field Ey=(300 V/m)sin ω(t−x/c). An electron is constrained to move along the Y-direction with a speed of 2.0×107 m/s. Find the maximum electric force and the maximum magnetic force on the electron respectively.

A
4.8×1017 N,3.2×1018 N
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
9.6×1017 N,6.4×1018 N
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
2.4×1017 N,1.6×1018 N
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
3.6×1017 N,2.5×1018 N
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

## The correct option is A 4.8×10−17 N,3.2×10−18 NGiven, Ey=300 sin ω[t−xc] E0=300 V/m Force due to electric field, FE=qE=eE=1.6×10−19×300=4.8×10−17 N As we know that, B0=E0c where, B0 = Max. magnetic field E0 = Max. electric field c= speed of light It is given that, E0=300 So, Bo=3003×108=10−6T Now, maximum magnetic force = qvB0 =1.6×10−19×2×107×10−6 =3.2×10−18 N

Suggest Corrections
2
Join BYJU'S Learning Program
Select...
Related Videos
Electromagnetic Spectrum
PHYSICS
Watch in App
Join BYJU'S Learning Program
Select...