Question

An ion source is producing 6Li ions, which have charge +e and mass 9.99 × 10-27 kg. The ions are accelerated by a potential difference of 9.2 kV and pass horizontally into a region in which there is a uniform vertical magnetic field of magnitude B = 0.99 T. Calculate the strength of the smallest electric field, to be set up over the same region, that will allow the 6Li ions to pass through undeflected.

Answer 1

`5.38\cdot 10^5 V/m`

Step-by-step explanation:

At first, the 6Li ions are accelerated by the potential difference, so their gain in kinetic energy is equal to the change in electric potential energy; so we can write:

`q\Delta V=(1)/(2)mv^2`

where

`q=+e=+1.6\cdot 10^(-19)C` is the charge of one 6Li ion

`\Delta V=9.2 kV=9200 V` is the potential difference through which they are accelerated

`m=9.99\cdot 10^(-27)kg` is the mass of each ion

v is the final speed reached by the ions

Solving for v, we find:

`v=\sqrt{(2q\Delta V)/(m)}=\sqrt{(2(1.6\cdot 10^(-19))(9200))/(9.99\cdot 10^(-27))}=5.43\cdot 10^5 m/s`

After that, the ions pass into a region with a uniform magnetic field of strength

`B=0.99 T`

The magnetic field exerts a force perpendicular to the direction of motion of the ions, and this force is given by

`F=qvB`

In order to make the ions passing through undeflected, there should be an electric force balancing this magnetic force. The electric force is given by

`F=qE`

where E is the strength of the electric field.

Since the two forces must be balanced,

`qE=qvB`

From which we get

`E=vB`

So the strength of the electric field must be

`E=(5.43\cdot 10^5)(0.99)=5.38\cdot 10^5 V/m`