JEE Main · 2022 · 27 Jun 2022 · Shift 2 · Magnetic Effects of Current

A deuteron and a proton moving with equal kinetic energy enter into to a uniform magnetic field at right angle to the field. If

r_{d}

and

r_{p}

are the radii of their circular paths respectively, then the ratio

\frac{r _{d}}{r _{p}}

will be

x : 1

where x is ______.

Answer: 2

Solution

The radius of circular motion in a magnetic field is:

r = \frac{m v}{q B}

For deuteron (mass

2 m_{p}

, charge

e

r_{d} = \frac{2 m _{p} v _{d}}{e B}

For proton (mass

m_{p}

, charge

e

r_{p} = \frac{m _{p} v _{p}}{e B}

The ratio is:

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

Since kinetic energies are equal:

\frac{1}{2} (2 m_{p}) v_{d}^{2} = \frac{1}{2} m_{p} v_{p}^{2}

2 v_{d}^{2} = v_{p}^{2}

v_{p} = 2 v_{d}

Therefore:

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

x = 2

JEE Main · 2022 · 27 Jun 2022 · Shift 2 · Integer answer