Question

A particle of charge Q and mass M moves in a circular path of radius R in a uniform magnetic field of magnitude B. The same particle now moves with the same speed in a circular path of same radius R in the space between the cylindrical electrodes of the cylindrical capacitor. The radius of the inner electrode is $\dfrac{R}{2}$ while that of the outer electrode is $\dfrac{{3R}}{2}$. Find the potential difference between the capacitor electrodes.

Answer 1

The radius of a particle performing circular motion in a magnetic field is $r = \dfrac{{mv}}{{QB}}$ . Find the time period taken to perform the circular motion. Potential difference between the capacitor electrodes are given as:
$V = \dfrac{Q}{{2\pi l{E_0}}}\ln \left( {\dfrac{b}{a}} \right)$

Formula used:
$r = \dfrac{{mv}}{{QB}}$
Where $m$ is the mass of the particle; $v$ is the velocity of the particle; $Q,B$ Have the same meanings as mentioned in the question.
$V = \dfrac{Q}{{2\pi l{\varepsilon _0}}}\ln \left( {\dfrac{b}{a}} \right)$
Where $V$is the potential difference; $b,a$ is the inner and outer radius of the electrode.

Complete step by step answer:
The charge on the given particle is $Q$. The mass of the given particle is $m$. The radius of the particle in the magnetic field is $R$. The initial magnetic field is given to be $B$.
The radius of the inner electrode $= \dfrac{R}{2}$
The radius of the outer electrode $= \dfrac{{3R}}{2}$
The particle then moves in the space between the cylindrical electrodes.
Let’s calculate the time period of the particle in the magnetic field, this will be given as follows:
$T = \dfrac{{2\pi }}{\omega }$
Where $\omega$ is the angular velocity.
The angular velocity is given as:
$\omega = \dfrac{v}{{\dfrac{{mv}}{{QB}}}}$
Therefore, the time period will be
$T = \dfrac{{2\pi m}}{{QB}}$
The potential difference is given as:
$V = \dfrac{Q}{{2\pi l{\varepsilon _0}}}\ln \left( {\dfrac{b}{a}} \right)$
$l = R$ , the radius of the circular motion is $R$
Here ${\varepsilon _0}$ is known as permittivity of free space and has constant value.
Substituting the values we will have:
$V = \dfrac{Q}{{2\pi R{\varepsilon _0}}}\ln \left( {\dfrac{{\dfrac{{3R}}{2}}}{{\dfrac{R}{2}}}} \right)$
$\therefore V = \dfrac{Q}{{2\pi R{\varepsilon _0}}}\ln \left( 3 \right)$

Therefore the required potential difference is $\dfrac{Q}{{2\pi R{\varepsilon _0}}}\ln \left( 3 \right)$.

Note: The final answer can be converted into many forms depending as per requirements. If the question was multiple choice based then we could also convert the final answer in terms of the magnetic field. Remember that magnetic field is the region around a magnetic material or a moving electric charge within which the force of magnetism acts.