Chapter 16, Problem 21A

Electrons are fired at different speeds through a magnetic field and are bent from their straight-line paths to hit the detector at the points shown. Rank the speeds of the electrons from highest to lowest.

To determine

To Rank: $v_A$ , $v_B$ and $v_C$ , where $v_A$ , $v_B$ and $v_C$ are the speed of electrons that hit at the point A, B and C respectively.

Answer to Problem 21A

  $v_A>v_B>v_C$

Explanation of Solution

Introduction: Force, $\overrightarrow{F}$ on a charged particle with charge e, moving with a velocity $\overrightarrow{v}$ in a magnetic field $\overrightarrow{B}$ is $\overrightarrow{F}=e(\overrightarrow{v}\times \overrightarrow{B})$

Deflection of charged particles will be greater when the force due to the magnetic field is greater and force will be greater when speed v will be greater.

Here all the three electron beams are in the same magnetic field $\overrightarrow{B}$ and all the electrons have the charge e.

Electron A have the greatest deflection, then electron B and then electron C has the lowest deflection.

Then the forces for the electrons for A, B, C are respectively.

  ${\overrightarrow{F}}_A=e({\overrightarrow{v}}_A\times \overrightarrow{B})$

  ${\overrightarrow{F}}_B=e({\overrightarrow{v}}_B\times \overrightarrow{B})$

  ${\overrightarrow{F}}_C=e({\overrightarrow{v}}_C\times \overrightarrow{B})$

From the order of deflection, the order of force is: $F_A>F_B>F_C$ .

So from the above the order, of speeds of electrons must be $v_A>v_B>v_C$ .