JEE PYQ: Magnetic Effects Of Current Question 51
Question 51 - 2019 (12 Jan 2019 Shift 2)
A proton and an $\alpha$-particle (with their masses in the ratio of $1:4$ and charges in the ratio $1:2$) are accelerated from rest through a potential difference $V$. If a uniform magnetic field ($B$) is set up perpendicular to their velocities, the ratio of the radii $r_p : r_\alpha$ of the circular paths described by them will be:
(1) $1:\sqrt{2}$
(2) $1:2$
(3) $1:3$
(4) $1:\sqrt{3}$
Show Answer
Answer: (1) $1:\sqrt{2}$
Solution
$r = \frac{1}{B}\sqrt{\frac{2mV}{q}}$. $\frac{r_p}{r_\alpha} = \sqrt{\frac{m_p q_\alpha}{q_p m_\alpha}} = \sqrt{\frac{1 \times 2}{1 \times 4}} = \frac{1}{\sqrt{2}}$.
BETA
