Question

A particle of mass 10 g and charge 80 mC moves through a uniform magnetic field,in a region where the free-fall acceleration is . The velocity of the particle is a constant , which is perpendicular to the magnetic field. What, then, is the magnetic field?

Answer 1

Answer:

$B=1.223\frac{T\cdot m}{s}\frac{1}{v}$

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:

$F_m=F_g$

Here $F_m=qvBsin\theta$ is the magnetic force and $F_g=mg$ is the gravitational force. The velocity of the particle is perpendicular to the magnetic field, so $\theta=90^\circ$. Replacing and solving for B:

$qvBsin(90^\circ)=mg\\B=\frac{mg}{qv}\\B=\frac{mg}{q}\frac{1}{v}\\B=\frac{(10*10^{-3}kg)(9.8\frac{m}{s^2})}{80*10^{-3}C}\frac{1}{v}\\B=1.223\frac{T\cdot m}{s}\frac{1}{v}$