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

Answer:

$5.38\cdot 10^5 V/m$

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:

$q\Delta V=\frac{1}{2}mv^2$

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:

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

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

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