Question

Launch 3 small particles from (-x0,0,0) with velocity u in the +x-direction thru a magnetic field, B=B0ẑ. Particle A has positive charge Q and mass m. Particle B has charge –Q and mass m. Particle C has charge Q and mass 2m. Mark (relatively) where on a screen (at x=0) each particle strikes.

Answer 1

Answer:

The particle A will strike on the screen to the right (in -y₀). The particle B will strike to the left of the screen (in y₀), at the same distance than particle A from the x-axis but in the opposite direction. The particle C will strike to the right of the screen (in -y₁), the same direction than particle A, but nearer to the x-axis (see attached image)

The exact positions in the screen are (the point [0,y,0]):

$Y_a=-y_0=\displaystyle -\frac{muB_0}{q}+\sqrt{\frac{m^2u^2B^2_0}{q^2}-x^2_0}$

$Y_b=y_0=\displaystyle \frac{muB_0}{q}-\sqrt{\frac{m^2u^2B^2_0}{q^2}-x^2_0}$

$Y_c=-y_1=\displaystyle -\frac{2muB_0}{q}+\sqrt{\frac{4m^2u^2B^2_0}{q^2}-x^2_0}$

Explanation:

The electric charges that move throw a region of space with a magnetic field will suffer a magnetic force (explain by Lorentz Force law). This force will force the particle to change direction but won't change its speed module. Therefore magnetic force act as a centripetal force.

The Lorentz Force law can be written as:

$\vec{F_B}=q\vec{v}\times \vec{B}$

For particle A:

$\vec{F_{Ba}}=qu\vec{x}\times B_0\vec{z}=quB_0(-\vec{y})$

For particle B:

$\vec{F_{Bb}}=-qu\vec{x}\times B_0\vec{z}=quB_0(\vec{y})$

For particle C:

$\vec{F_{Bc}}=qu\vec{x}\times B_0\vec{z}=quB_0(-\vec{y})=\vec{F_{Ba}}$

The force applied in each particle in the module is the same as you can see. Nevertheless, their directions are not. In the case of particles A and C, the force has a negative direction in the y-axis while in case B has a positive direction in the y-axis.

Knowing that the magnetic force is a centripetal force, we can find the radius of curvature:

$|F_B|=\displaystyle m\frac{v^2}{r}$

For particle A:

$|F_{Ba}|=\displaystyle m\frac{v^2}{r}= quB_0 \rightarrow r=\frac{muB_0}{q}$

For particle B:

$|F_{Bb}|=\displaystyle m\frac{v^2}{r}= quB_0 \rightarrow r=\frac{muB_0}{q}$

For particle C:

$|F_{Bc}|=\displaystyle 2m\frac{v^2}{r}= quB_0 \rightarrow r=\frac{2muB_0}{q}$

Now we can obtain the exact point in the screen where the particle will strike. We can see than particle A and C are affected by the same force (same module and direction), but the radius of curvature of particle C is twice the one of particle A. Therefore the particle C will strike nearer to the x-axis than particle A.

In each case we can use Pythagoras Theorem to determine the point Y where the particles strike:

$y+L=r$   and in the triangle form   $L^2+x_0^2=r^2$

Therefore:

$y=r-\sqrt{r^2-L^2}$