After being accelerated through a potential difference of 12 kV , a...

The kinetic energy gained by the charged particle when accelerated through a potential difference $V$ is given by:

$\dfrac{1}{2}mv^2 = qV$

$\Rightarrow \dfrac{m}{q} = \dfrac{2V}{v^2}$

When the particle is injected perpendicularly into a magnetic field $B$, the radius $r$ of the circular path is:

$r = \dfrac{mv}{qB}$

Substituting the value of $\dfrac{m}{q}$ into the expression for $r$:

$r = \left(\dfrac{2V}{v^2}\right) \dfrac{v}{B} = \dfrac{2V}{vB}$

Given $V = 12 \times 10^3 \text{ V}$, $v = 1.0 \times 10^6 \text{ m s}^{-1}$, and $B = 0.2 \text{ T}$, we get:

$r = \dfrac{2 \times 12 \times 10^3}{1.0 \times 10^6 \times 0.2}$

$r = \dfrac{24 \times 10^3}{2 \times 10^5}$

$r = 12 \times 10^{-2} \text{ m} = 12 \text{ cm}$