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 paths or radii $R_1$ and $R_2$ respectively. The ratio of mass of X to that of Y is equal to :

• A. ${\left(\frac{R_1}{R_2}\right)}^2$
• B. $\left(\frac{R_1}{R_2}\right)$
• C. ${\left(\frac{R_1}{R_2}\right)}^{1/2}$
• D. $\frac{R_2}{R_1}$

Answer 1

The correct option is A ${\left(\frac{R_1}{R_2}\right)}^2$

Force on a charged particle q having velocity $v$ in the magnetic field(B) is = $F_B$ = q$v$B
Centripetal force on the particle moving in a circular path = $F_C$ = $\frac{m{v}^2}{R}$
Now for particle to remain in circular path = $F_B=F_C$
q$v$B = $\frac{m{v}^2}{R}$
q$v_1$B = $\frac{m_1{v_1}^2}{R_1}$ (For particle 1) $\Rightarrow$ qB = $\frac{m_1{v}_1}{R_1}$ .......eq.(1)
q$v_2$B = $\frac{m_2{v_2}^2}{R_2}$ (For particle 2) $\Rightarrow$ qB = $\frac{m_2{v}_2}{R_2}$ .......eq.(2)
By equating eq.(1) & (2) $\Rightarrow$ $\frac{m_1{v}_1}{R_1}$ = $\frac{m_2{v}_2}{R_2}$
$\left(\frac{m_1}{m_2}\right)=\left(\frac{v_2}{v_1}\right)\times \left(\frac{R_1}{R_2}\right)$ ........eq.(3)
By accelerating the charged particle q with voltage (V), work done by the particle(W) = qV
W = $\Delta$K (work energy theorem)
qV = $\frac{1}{2}m{v}^2$ - 0
$\Rightarrow \frac{1}{2}{m}_1{v_1}^2=qV=\frac{1}{2}{m}_2{v_2}^2$
$\Rightarrow m_1{v_1}^2=m_2{v_2}^2$
$\left(\frac{v_2}{v_1}\right)=(\frac{m_1}{m_2}{)}^{\frac{1}{2}}$ ..........eq.(4)
Now from eq.(3) & (4) $\Rightarrow$ $\left(\frac{m_1}{m_2}\right)=\left(\frac{m_1}{m_2}{)}^{\frac{1}{2}}\right(\frac{R_1}{R_2})$
$(\frac{m_1}{m_2}{)}^{\frac{1}{2}}$ = $\left(\frac{R_1}{R_2}\right)$
$\left(\frac{m_1}{m_2}\right)$ = $(\frac{R_1}{R_2}{)}^2$