1. In a uniform magnetic field, a charged particle moving perpendicular to the field follows a circular path of radius

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

where $m$ is mass, $v$ is speed, $q$ is charge, and $B$ is magnetic field.

2. Since $m$, $q$, and $B$ remain constant, we get

$r \propto v$

3. The kinetic energy of the particle is

$K = \frac{1}{2}mv^2$

So,

$K \propto v^2$

4. It is given that the new kinetic energy becomes $4$ times the initial kinetic energy:

$K_2 = 4K_1$

Thus,

$\frac{v_2^2}{v_1^2} = 4 \implies \frac{v_2}{v_1} = 2$

5. Since radius is directly proportional to speed,

$\frac{r_2}{r_1} = \frac{v_2}{v_1} = 2$

Therefore, the ratio of new radius to original radius is

$r_2 : r_1 = 2:1$

6. Checking options:

• A: $1:1$ — incorrect
• B: $1:2$ — incorrect
• C: $2:1$ — correct
• D: $1:4$ — incorrect