Answer

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**Hint:**To solve this question, we need to look for using the concept of centripetal force and apply it along with the concept of force on a charge. The concept of conservation of energy is also useful in this problem.

**Complete step by step answer:**

It is given that the charged particle is accelerated through a potential difference of $24\times{10^3}\,V$.And as a result it acquires a speed of $2\times {10^6}\,m/s$. Now from the question we can easily understand that the centripetal force will be equal to the force on the charge and we can use this equation for further parts of the question. Therefore,

$\dfrac{m{{v}^{2}}}{r}=qVB$ ,

simplifying this equation we get the following

$r=\dfrac{mv}{qB}$..........................$\left( 1 \right)$

Moving ahead, with the help of conservation of energy theorem, we can say that the kinetic energy of the charge initially will be equal to the potential energy of the charge after it injects into the magnetic field. Therefore, according to conservation of energy theorem,

$\dfrac{1}{2}m{{v}^{2}}=q{{V}_{B}}$

After simplifying, we get that

$\dfrac{1}{q}=\dfrac{2{{V}_{B}}}{m{{v}^{2}}}$.................$\left( 2 \right)$

Now using both of the equation and putting the value of $\dfrac{1}{q}$ in the first equation

$r=\dfrac{2{{V}_{B}}}{m{{v}^{2}}}\times \dfrac{mv}{B}$

we can get that

$r=\dfrac{2{{V}_{B}}}{BV}$

Now, putting the values we get the answer

\[r=\dfrac{\left( 2\times 24 \times{10^3}\right)}{\left( 2\times 0.2\times {{10}^{6}} \right)}\\

\Rightarrow r=\dfrac{48\times{10^3}}{0.4\times{10^6}}\\

\Rightarrow r=0.120\,m\\

\therefore r=12\,cm\]

**Therefore the radius of the circle described in the magnetic field is $12\,cm$.**

**Note:**The question says that the charged particle is inserted perpendicularly, so we solve the question as above, but if the question had any other conditions like the charge is inserted parallel or at a given angle then the solution would have changed slightly.

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