Question #312730

A particle of charge 4.25 x 10-⁹C moves at 3.0 x 10 ⁴ m/s at 60° up the x-y plane through a uniform magnetic field of


5.0 x 10-³T directed along the +z - axis. Find the magnetic force on a particle.

1
Expert's answer
2022-03-16T18:28:47-0400

The magnetic force on a particle is given by

F=qvBsinθ=4.25109C3.0104m/s5.0103Tsin30=3.4107NF=qvB\sin\theta\\ ={\rm 4.25*10^{-9}\:C*3.0*10^4\: m/s}\\ *5.0*\rm 10^{-3}\: T*\sin30^\circ=3.4*10^{-7}\: N


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS