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

Particles A, B, and C, with different charges and masses, are launched through a magnetic field. The particles will strike a screen at x=0 when their circular paths intersect the screen. The radius of each particle's path can be determined using the equation r = mv/(|q|B).

Step-by-step explanation:

The three particles, A, B, and C, are launched with velocity u in the +x direction through a magnetic field B=B0ẑ. Particle A has a positive charge Q and mass m, Particle B has a charge -Q and mass m, and Particle C has a charge Q and mass 2m. To determine where each particle strikes on a screen at x=0, we need to consider the motion of charged particles in a magnetic field. Charged particles experience a magnetic force when moving in a magnetic field, given by the equation F = qvB, where q is the charge, v is the velocity, and B is the magnetic field. The magnetic force acts as a centripetal force, causing the particles to move in circular paths. The radius of the path can be calculated using the equation r = mv/(|q|B). Particle A, with charge Q and mass m, will have a radius of rA = mu/(Q|B|), Particle B, with charge -Q and mass m, will have a radius of rB = mu/(|Q||B|), and Particle C, with charge Q and mass 2m, will have a radius of rC = 2mu/(Q|B|). The particles will strike the screen at x=0 when their paths intersect the screen at that location.

Answer 2

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 -(muB_0)/(q)+\sqrt{(m^2u^2B^2_0)/(q^2)-x^2_0}`

`Y_b=y_0=\displaystyle (muB_0)/(q)-\sqrt{(m^2u^2B^2_0)/(q^2)-x^2_0}`

`Y_c=-y_1=\displaystyle -(2muB_0)/(q)+\sqrt{(4m^2u^2B^2_0)/(q^2)-x^2_0}`

Step-by-step 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}* \vec{B}`

For particle A:

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

For particle B:

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

For particle C:

`\vec{F_(Bc)}=qu\vec{x}* 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(v^2)/(r)`

For particle A:

`|F_(Ba)|=\displaystyle m(v^2)/(r)= quB_0 \rightarrow r=(muB_0)/(q)`

For particle B:

`|F_(Bb)|=\displaystyle m(v^2)/(r)= quB_0 \rightarrow r=(muB_0)/(q)`

For particle C:

`|F_(Bc)|=\displaystyle 2m(v^2)/(r)= quB_0 \rightarrow r=(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-√(r^2-L^2)`