JEE Main · 2019 · 12 Jan 2019 · Shift 1 · Magnetic Effects of Current

A proton and an

α

-particle (with their masses in the ratio of

1 : 4

and charges in the ratio of

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_{α}

of the circular paths described by them will be:

(A) $1 : 2$ ✓ Correct
(B) $1 : 2$
(C) $1 : 3$
(D) $1 : 3$

Answer: (A)

Solution

After acceleration through

V

a charge

q

gains kinetic energy

q V

, so its momentum is

p = 2 m q V

and its radius in the field is

r = \frac{p}{q B} = \frac{2 m q V}{q B} = \frac{1}{B} \frac{2 mV}{q}

. The same

V

and

B

act on both particles, so

r \propto \frac{m}{q}

. For the proton

\frac{m}{q}

stays

\frac{m}{q}

, while for the

α

-particle it is

\frac{4 m}{2 q} = \frac{2 m}{q}

, so

\frac{r _{p}}{r _{α}} = \frac{m / q}{2 m / q} = \frac{1}{2}

, that is

1 : 2

JEE Main · 2019 · 12 Jan 2019 · Shift 1 · Single correct option

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JEE Main · 2019 · 12 Jan 2019 · Shift 1 · Magnetic Effects of Current

Solution

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