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**Hint:**This is a problem of cylindrical capacitor. Break the question into segments and then solve. Write the formula for electric potential and field for a cylindrical capacitor. Then apply the concept of centripetal and magnetic force on a charge along with the magnetic field and form a relation between the three.

**Complete step by step solution:**

Step 1: Write the formula electric field and potential for a cylindrical capacitor.

The potential is given as:

$\Delta V = - E. dr$;

Take integration on both the sides:

$\int {\Delta V} = - \int\limits_a^b {E.dr} $;

$\int {\Delta V} = - \int\limits_a^b {\dfrac{\lambda }{{2\pi {\varepsilon _o}r}}.dr} $; ….(Here: $E = \dfrac{\lambda }{{2\pi {\varepsilon _o}r}}$)

Take the constants out of the integration:

$\int {\Delta V} = - \dfrac{\lambda }{{2\pi {\varepsilon _o}}}\int\limits_a^b {\dfrac{1}{r}.dr} $;

Solve the integration by applying the integration property:$\int {\dfrac{1}{r}dr} = \ln r$

\[V = - \dfrac{\lambda }{{2\pi {\varepsilon _o}}}\left[ {\ln r} \right]_a^b\];

Apply the logarithmic property: \[\left[ {\ln r} \right]_a^b = \left[ {\ln b - \ln a} \right]\], in the above equation

\[V = - \dfrac{\lambda }{{2\pi {\varepsilon _o}}}\left[ {\ln b - \ln a} \right]\];

Here, \[\left[ {\ln b - \ln a} \right] = \ln \dfrac{b}{a}\]:

\[V = - \dfrac{\lambda }{{2\pi {\varepsilon _o}}}\ln \dfrac{b}{a}\];

The Electric field is given as:

$E = \dfrac{\lambda }{{2\pi {\varepsilon _o}r}}$;

$E. r = \dfrac{\lambda }{{2\pi {\varepsilon _o}}}$;

Put E in place of $\dfrac{\lambda }{{2\pi {\varepsilon _o}}}$ in the equation for electric potential:

\[V = - E.r\ln \dfrac{b}{a}\];

Write the above equation in terms of electric potential;

\[E = \dfrac{{ - V}}{{r\ln \dfrac{b}{a}}}\];

Step 2: Now, we know the magnetic force on a charge particle:

$F = q. (v \times B)$;

We also know the centripetal force on a charge in a magnetic field

\[{F_c} = \dfrac{{m{v^2}}}{r}\];

Here the magnetic force supplies the centripetal force, so we can equate them together:

\[q. (v \times B) = \dfrac{{m{v^2}}}{r}\];

Now, electric field is equal to:

\[E = (v \times B)\];

Put the above relation in the equation between centripetal and magnetic force:

\[q. E = \dfrac{{m{v^2}}}{r}\];

Put the value of the electric field \[E = \dfrac{V}{{r\ln \dfrac{b}{a}}}\]. In the above relation

\[q. \dfrac{V}{{r\ln \dfrac{b}{a}}} = \dfrac{{m{v^2}}}{r}\];

Cancel out the common;

\[m{v^2} = q. \dfrac{V}{{\ln (b/a)}}\];

We have already established that the centripetal force ${F_c} = \dfrac{{m{v^2}}}{r}$ is equal to the magnetic force ${F_m} = qvb$:

\[\dfrac{{m{v^2}}}{r} = qvB\];

Cancel out the common variable;

\[\dfrac{{mv}}{r} = qB\];

Take “r” to RHS we have:

\[mv = qBr\];

Put this relation in the equation \[m{v^2} = q. \dfrac{V}{{\ln (b/a)}}\]:

\[(qBr). v = q. \dfrac{V}{{\ln (b/a)}}\];

Cancel out the common variable on both the sides:

\[(Br). v = \dfrac{V}{{\ln (b/a)}}\];

Solve for the velocity,

\[v = \dfrac{V}{{Br\ln (b/a)}}\];

\[v = \dfrac{V}{{Br\ln (b/a)}}\];

Step 3: Find specific charge (q/m).

We know that:

$mv = qBr$;

Write the above equation in terms of q/m.

$\dfrac{q}{m} = \dfrac{{Br}}{v}$;

Put the value of \[v = \dfrac{V}{{Br\ln (b/a)}}\] in the above relation and solve,

$\dfrac{q}{m} = \dfrac{{Br}}{{\dfrac{V}{{Br\ln (b/a)}}}}$;

Solve,

$\dfrac{q}{m} = \dfrac{V}{{Br \times Br\ln (b/a)}}$;

$\dfrac{q}{m} = \dfrac{V}{{B{r^2}\ln (b/a)}}$;

**The velocity of the particle and its specific charge q/m is \[v = \dfrac{V}{{Br\ln (b/a)}}\]and $\dfrac{q}{m} = \dfrac{V}{{B{r^2}\ln (b/a)}}$.**

**Note:**It is a very lengthy process, be careful while formulating relations and writing the formulas. Go step by step. First write the potential and electric field and establish a relation between them. Then apply centrifugal force and magnetic force and equate them together. Then apply a magnetic field and find out the velocity and then the specific charge.

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