Download the app

Questions  

A particle of specific charge q/m=πC / kg is projected from the origin towards positive x-axis with a velocity of 10 m/sin a uniform magnetic field B=-2k^Tesla. The velocity V of the particle after time t= 1/6 s will be

a
(5i^+53j^)m/s
b
10j^ m/s
c
(53i^-5j^)m/s
d
-10j^ m/s

detailed solution

Correct option is A

Timle period T=2πmqB=1 s  Thus, the particle will be at P after t=16 s ∴  V→=10(cos60i^+sin60j^)=5(i^+3j^)m/s

Talk to our academic expert!

+91

Are you a Sri Chaitanya student?


Similar Questions

What is the value of B (in ×10-8 T ) that can be set up at the equator to permit a proton of speed 107 m/s to circulate around the earth? R=6.4×106 m,mP=1.67×10-27 kg


phone icon
whats app icon