A proton and a deutron ( $\mathrm{q}=+\mathrm{e}, m=2.0 \mathrm{u}$ ) having same kinetic energies enter a region of uniform magnetic field $\overrightarrow{\mathrm{B}}$, moving perpendicular to $\overrightarrow{\mathrm{B}}$. The ratio of the radius $r_{d}$ of deutron path to the radius $r_{p}$ of the proton path is :

JEE Main · 2024 · 09 Apr 2024 · Shift 2 · Magnetic Effects of Current

A proton and a deutron (

q = + e, m = 2.0 u

) having same kinetic energies enter a region of uniform magnetic field

B

, moving perpendicular to

B

. The ratio of the radius

r_{d}

of deutron path to the radius

r_{p}

of the proton path is :

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

Answer: (C)

Solution

For a charged particle moving in a magnetic field with kinetic energy

K

, the radius is

r = \frac{2 m K}{q B}

For the same

K

and

B

r \propto \frac{m}{q}

.

Deuteron:

m_{d} = 2 m_{p}, q_{d} = q_{p} = e

. So

\frac{r _{d}}{r _{p}} = \frac{m _{d}}{m _{p}} = 2

i.e.

r_{d} : r_{p} = 2 : 1

JEE Main · 2024 · 09 Apr 2024 · Shift 2 · Single correct option

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