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labrat256 | (Level 2) Adjunct Educator

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We are looking to find the charge of a particle with a known mass, moving in a magnetic field with the same path as that of one with a known charge and different mass.

The equation that relates charge, magnetic field strength, velocity and force is the Lorenz equation:

`F = q(E+vxxB)`

where F is the force, E is the electric field (in this question assumed to be 0), v is the velocity and B is the strength of the magnetic field. It can be deduced that because particle `m_2`  is curving in an opposite direction to mass `m_1`, that the charges must be related by `q_1=-xq_2`  for some positive x. Knowing this, the direction of the velocity and be ignored and from now on, I will focus only on the magnitude of the vectors.

The equation which incorporates mass and force is Newton's second law:

`F=ma`

where m is the particle mass and a is the acceleration.

In circular motion, the acceleration is of a fixed intensity perpendicular to the velocity. The final equation we need is one which links the magnitude of the acceleration with the magnitude of the velocity of the particle movement and the length of the radius with which it moves around in the circle. This equation for constant circular motion is:

`a=v^2/R`

where R is the circle radius. This is explained in the third reference below.

So, we now need to put these equations together. Joining the first two equations, and taking into account the fact that E=0, we get:

`ma=qvB`

and by plugging in the third equation and cancelling v, we get:

`mv/R=qB`

Looking at the diagram, it can be seen that the two particles collide when particle `m_1`  has travelled 1/4 of the way around the circle anticlockwise whilst `m_2`  has travelled 3/4 of the way around the circle. From this we can deduce that `|v_2|=3|v_1|` .

We now have a fixed R, a fixed B, a known m for each of the particles, a known relationship between the two velocities and one known charge. By rearranging the equation so that q and m are on the same side, and by plugging all of this information in, for `m_1`  we get:

`q_1/m_1=v_1/(RB)`

and for `m_2`  we get:` `

`|q_2|/m_2=v_2/(RB)`

which as shown above is the same as:

`q_2/m_2=3v_1/(RB)`

Dropping the units for a moment, we know that `q_1=0.1`  and `m_1=0.2`  so we can say that 

`v_1/(RB) = 0.1/0.2 = 0.5`

We know that `m_2 = 0.4` and this can all be plugged into the equation for the second particle as:

`|q_2|/m_2=3v_1/(RB)rArr|q_2|/0.4=3xx0.5rArr|q_2|=3xx0.5xx0.4=0.6`

Therefore, plugging in the units again `q_2=-0.6C`  which is answer A on your sheet.

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