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.

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:

[tex]q\Delta V=\frac{1}{2}mv^2[/tex]

where

[tex]q=+e=+1.6\cdot 10^{-19}C[/tex] is the charge of one 6Li ion

[tex]\Delta V=9.2 kV=9200 V[/tex] is the potential difference through which they are accelerated

[tex]m=9.99\cdot 10^{-27}kg[/tex] is the mass of each ion

v is the final speed reached by the ions

Solving for v, we find:

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

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

[tex]B=0.99 T[/tex]

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

[tex]F=qvB[/tex]

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

[tex]F=qE[/tex]

where E is the strength of the electric field.

Since the two forces must be balanced,

[tex]qE=qvB[/tex]

From which we get

[tex]E=vB[/tex]

So the strength of the electric field must be

[tex]E=(5.43\cdot 10^5)(0.99)=5.38\cdot 10^5 V/m[/tex]