A deuteron and a proton moving with equal kinetic energy enter into to a uniform magnetic field at right angle to the field. If rd and rp are the radii of their circular paths respectively, then the ratio (rd/rp) will be √ x: 1 where x is

Question

A deuteron and a proton moving with equal kinetic energy enter into to a uniform magnetic field at right angle to the field. If rd and rp are the radii of their circular paths respectively, then the ratio (rd/rp) will be √ x: 1 where x is  (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/4ef2f98d9ef1f7b1f67bd2c5bdae164d-.png" /> R =( mv / q B ) R D =((2 m P ) v D / eB ) R P =(( m P ) v P / e B ) ( R D / R P )=(2 v D / v P )=(2 v D /√2 v D )=(√2/1) (1/2)(2 mp ) v D 2=(1/2) m P ⋅ v P 2 √2 v D = v P x =2

Q. 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 _____

$R = \frac{m v}{q _{B}}$
$R_{D} = \frac{( 2 m _{P} ) v _{D}}{e B}$
$R_{P} = \frac{( m _{P} ) v _{P}}{e B}$
$\frac{R _{D}}{R _{P}} = \frac{2 v _{D}}{v _{P}} = \frac{2 v _{D}}{2 v _{D}} = \frac{2}{1}$
$\frac{1}{2} (2 m p) v_{D}^{2} = \frac{1}{2} m_{P} \cdot v_{P}^{2}$
$2 v_{D} = v_{P}$
$x = 2$

Q. A deuteron and a proton moving with equal kinetic energy enter into to a uniform magnetic field at right angle to the field. If rd​ and rp​ are the radii of their circular paths respectively, then the ratio rp​rd​​ will be x​:1 where x is _____

R=qB​mv​ RD​=eB(2mP​)vD​​ RP​=eB(mP​)vP​​ RP​RD​​=vP​2vD​​=2​vD​2vD​​=12​​ 21​(2mp)vD2​=21​mP​⋅vP2​ 2​vD​=vP​ x=2

Q. 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 _____

$R = \frac{m v}{q _{B}}$
$R_{D} = \frac{( 2 m _{P} ) v _{D}}{e B}$
$R_{P} = \frac{( m _{P} ) v _{P}}{e B}$
$\frac{R _{D}}{R _{P}} = \frac{2 v _{D}}{v _{P}} = \frac{2 v _{D}}{2 v _{D}} = \frac{2}{1}$
$\frac{1}{2} (2 m p) v_{D}^{2} = \frac{1}{2} m_{P} \cdot v_{P}^{2}$
$2 v_{D} = v_{P}$
$x = 2$