Answer:
[tex]B=1.223\frac{T\cdot m}{s}\frac{1}{v}[/tex]
Explanation:
According to the first Newton's law, when a particle moves with constant velocity, the sum of forces on it is zero. So, we have:
[tex]F_m=F_g[/tex]
Here [tex]F_m=qvBsin\theta[/tex] is the magnetic force and [tex]F_g=mg[/tex] is the gravitational force. The velocity of the particle is perpendicular to the magnetic field, so [tex]\theta=90^\circ[/tex]. Replacing and solving for B:
[tex]qvBsin(90^\circ)=mg\\B=\frac{mg}{qv}\\B=\frac{mg}{q}\frac{1}{v}\\B=\frac{(10*10^{-3}kg)(9.8\frac{m}{s^2})}{80*10^{-3}C}\frac{1}{v}\\B=1.223\frac{T\cdot m}{s}\frac{1}{v}[/tex]
