An ion source is producing${}^{\mathbf{6}}\mathbf{Li}$ions, which have charge$\mathbf{+}\mathbf{e}$and mass$\mathbf{\text{9.99}}\mathbf{\times }{\mathbf{\text{10}}}^{\mathbf{-}\mathbf{27}}\mathbf{\text{kg}}$. The ions are accelerated by a potentialdifference of$\mathbf{\text{10kV}}$ and pass horizontally into a region in whichthere is a uniform vertical magnetic field of magnitude$\mathbf{B}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{2}\mathbf{\text{T}}$.Calculate the strength of the smallest electric field, to be set upover the same region that will allow the${}^{\mathbf{6}}\mathbf{Li}$ ions to pass throughun-deflected.

a) Mass of the ${}^{6}Li$ion, $m=9.99\times {10}^{-27}\text{kg}$

b) Charge of ${}^{6}Li$ion is $q=+e$.

c) Potential difference due to acceleration of ions, $V=10\text{kV}$

d) Magnitude of uniform vertical magnetic field, $B=1.2\text{T}$

For the required condition of un-deflected the lithium-ion, the electric force acting on the body needs to balance with the Lorentz force acting on the ion due to the presence of the uniform magnetic field. Again, the electric force is given by the electric field that is further given by the accelerated potential difference.

Formulae:

The electric force acting on a body due to its charge is as follows:

$F_e=qE$ ….. (i)

The Lorentz force acting on the body due to uniform magnetic field is as follows:

$F_B=qvB\sin \theta$ …… (ii)

The electric potential of a charge due to accelerating potential difference is as follows:

$E=qV$ …… (iii)

The kinetic energy of a body in motion is as follows:

$K=\frac{1}{2}m{v}^2$ ……. (iv)

Now, we are given that the magnetic field is perpendicular to the velocity.

Thus, the magnetic force using equation (ii) can be given as:

$$\begin{gathered} F_B=qvB\sin {90}^0 \\ =qvB \end{gathered}$$ …… (a)

Now, again the lithium ion is being accelerated by a potential difference and thus according to energy conservation, we get the speed of the lithium ion using equations (iii) and (iv) as follows:

$\begin{array}{rcl}\frac{1}{2}m{v}^2& =& eV\\ v& =& \sqrt{\frac{2eV}{m}}\end{array}$ ….. (b)

Now, using value of equation (b) in equation (a) and then equating that to electric force of equation (i) for the given condition of un-deflection as follows:

$$\begin{gathered} q\left(\sqrt{\frac{2eV}{m}}\right)B=qE \\ E=B\sqrt{\frac{2eV}{m}} \\ E=\left(1.2\text{T}\right)\sqrt{\frac{2\left(1.6\times {10}^{-19}\text{C}\right)\left(10\times {10}^3\text{V}\right)}{\left(9.99\times {10}^{-27}\text{kg}\right)}} \\ E=6.8\times {10}^5\frac{\text{V}}{\text{m}} \end{gathered}$$

Hence, the value of the electric field is $6.8\times {10}^5\frac{\text{V}}{\text{m}}$.