Question

Two particles X and Y having equal charges, after being accelerated through the same potential difference, enter a region of uniform magnetic field and describe circular path of radius $r_1$ and $r_2$ respectively. The ratio of mass of X to that of Y is ___________.

• A. $\sqrt{\left(r_1/r_2\right)}$
• B. $\left(r_2/r_1\right)$
• C. ${\left(r_1/r_2\right)}^2$
• D. $r_1/{r_2}^2$

Answer 1

The correct option is C

${\left(r_1/r_2\right)}^2$

Explanation

Correct option is C.

Concept used:

• The electromagnetic force on a charged particle is called as Lorentz force.
• When a charged particle of mass m, moves perpendicularly in an uniform magnetic field (B), direction of the particle changes but velocity remains constant as magnetic force is always perpendicular to the velocity.
• This magnetic force provides centripetal force and keeps the charge moving in circular path.

Step 1: Given data:

• Charges on X and Y are $q$and $q$ respectively.
• Potential difference is $V$.
• Magnetic field is $B$
• Radii of two circular tracks are $r_1$ and $r_2$ respectively.
• Let the mass of X and Y are $m_A$ and $m_B$ respectively.

Step 2: Formula used:

• The force on the charge, $\overrightarrow{F}=q\left(\overrightarrow{v}\times \overrightarrow{B}\right)$, where $q$ is charge on the particle, $v$ is its velocity and $B$ is the magnetic field.
• Centripetal force $=\frac{m{v}^2}{r}$, where $m$ is the mass, $v$ is its velocity and $r$ is the radius of the circular path.
• Acquired potential energy of a charged particle when passed through a potential $V$is $qV$

Step 3: Calculating the ratio of mass of X and Y:

$$\begin{gathered} Magneticforce=Centripetalforce \\ qvB=\frac{m{v}^2}{r} \\ Hencer=\frac{mv}{qB} \\ v=\frac{rqB}{m}---\left(1\right) \end{gathered}$$

Given that particles are accelerated through same potential difference, therefore they have same potential energy (P. E) and kinetic energy (K. E).

$$\begin{gathered} K.E=P.E \\ \frac{1}{2}m{v}^2=qV \\ v=\sqrt{\frac{2qV}{m}} \end{gathered}$$

Equation 1 can be written as

$\sqrt{\frac{2qv}{m}}=\frac{rqB}{m}$

From the above relation mass of the particle and radii are related as

$$\begin{gathered} \sqrt{m}\alpha r \\ or\sqrt{\frac{m_1}{m_2}}=\frac{r_1}{r_2} \\ \frac{m_1}{m_2}={\left(\frac{r_1}{r_2}\right)}^2 \end{gathered}$$

Hence option C is correct.