Answer

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**Hint:**Use the formula of the radius of the curved path, substitute the formula of the magnetic field in that. Rearrange the obtained equation and substitute the value of the radius in it to find the value of the velocity of the particle that does not strike the solenoid.

**Formula used:**

(1) The radius of the curved path is given by

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

Where $r$ is the radius of the circular path, $m$ is the mass of the particle, $v$ is the velocity of the particle, $B$ is the magnetic field produced by the solenoid and $q$ is the charge of the particle.

(2) The magnetic field of the solenoid is given by

$B = {\mu _0}ni$

Where ${\mu _0}$ is the magnetic permeability of free space, $n$ is the number of turns per unit length of the solenoid and $i$ is the current through the solenoid.

**Complete step by step solution:**

It is given that the

Number of turns in the solenoid is $n$

Radius of the solenoid is $r$

The current of the solenoid is $i$

The particle that travels perpendicular to the axis possesses the charge $q$ and mass $m$.

Using the formula of the radius of the curved path,

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

Substituting the formula of the magnetic field in the above step, we get

$\Rightarrow r = \dfrac{{mv}}{{{\mu _0}niq}}$

Rearranging the above formula, to obtain the value of the velocity of the particle

$\Rightarrow v = \dfrac{{{\mu _0}niqr}}{m}$

The radius of the curved path will be $\dfrac{r}{2}$ to obtain the speed that the particle does not strike the solenoid

$\Rightarrow v = \dfrac{{{\mu _0}niqr}}{{2m}}$

Hence the maximum velocity of the particle that does not strike the solenoid is $\dfrac{{{\mu _0}niqr}}{{2m}}$ .

**Thus the option (A) is correct.**

**Note:**In this problem, the magnetic field is produced due to the passing of the current through it. The radius of the circle is taken as its half, this is because if the radius is taken as full, the particle may strike the solenoid. Hence the radius of the circle taken as $\dfrac{r}{2}$.

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