Question

A particle of mass 11.9 g and charge 89.4 muC moves through a uniform magnetic field, in a region where the free-fall acceleration a -9.8 j m/s^2 without falling. The velocity of the particle is a constant 23.6 t km/s, which is perpendicular to the magnetic field. What, then, is the magnetic field? Number (i + j + k) Units

Answer 1

mass of the particle, m=11.9g

charge q=89.4 uC

acceleration, a=-9.8 j m/sec^2

velocity of the partivle, V=23.6 i km/sec

===>

use,

F=q*(VXB)

m*a =89.4*10^-6*(vx i+ Vy j+ Vz k)X((Bx)i+(By)j+(Bz)k)

11.9*10^-3*(-9.8)j=89.4*10^-6*(23.6*10^3i+0j+0k)X((Bx)i+(By)j+(Bz)k)

11.9*10^-3*(-9.8)j=89.4*10^-6*((23.6*10^3By)k-(23.6*Bz)j)

11.9*10^-3*(-9.8)j=(89.4*10^-6)*(23.6*10^3By)k- (89.4*10^-6)*(23.6*10^3Bz)j

===>

comparing both sides,

Bx=0,

-(89.4*10^-6)*(23.6*10^3*Bz)= 11.9*10^-3*(-9.8)

(89.4*10^-6)*(23.6*10^3*Bz)= 11.9*9.8*10^-3

==> Bz=0.055 T

By=0

====>

magnetic field, B=(Bx)i+(By)j+(Bz)k

B=0i+0j+(0.055)k T