An alpha-particle and a proton moving with th | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The radius $r$ of a circular path of a charged particle moving in a uniform magnetic field $B$ is given by $r = \frac{mv}{qB} = \frac{p}{qB}$.Sinc

Vedclass · Accepted Answer

$1: 1$

Vedclass · Answer

(B) The radius $r$ of a circular path of a charged particle moving in a uniform magnetic field $B$ is given by $r = \frac{mv}{qB} = \frac{p}{qB}$.Since kinetic energy $K = \frac{p^2}{2m}$,the momentum $p = \sqrt{2mK}$.Substituting this into the radius formula,we get $r = \frac{\sqrt{2mK}}{qB}$.For an $\alpha$-particle,mass $m_{\alpha} = 4m_p$ and charge $q_{\alpha} = 2e$. For a proton,mass $m_p$ and charge $q_p = e$.Given that both have the same kinetic energy $K$ and enter the same magnetic field $B$,the ratio of the radii is:$\frac{r_{\alpha}}{r_p} = \frac{\sqrt{m_{\alpha}} / q_{\alpha}}{\sqrt{m_p} / q_p} = \frac{\sqrt{4m_p} / 2e}{\sqrt{m_p} / e} = \frac{2\sqrt{m_p} / 2e}{\sqrt{m_p} / e} = \frac{1}{1}$.Thus,the ratio is $1: 1$.

An $\alpha$-particle and a proton moving with the same kinetic energy enter a region of uniform magnetic field at right angles to the field. The ratio of the radii of the paths of $\alpha$-particle to that of the proton is

Similar Questions

An electron is moving along the positive $x$-axis. To make it move in an anticlockwise circular path in the $x-y$ plane,a magnetic field must be applied:

An electron is projected along the axis of a circular conductor carrying some current. The electron will experience a force:

An electron with energy $880 \,eV$ enters a uniform magnetic field of induction $2.5 \times 10^{-3} \,T$. The radius of the circular path will approximately be:

$A$ charge $Q$ moves parallel to a very long straight wire carrying a current $I$ as shown. The force on the charge is

An $\alpha$-particle travels in a circular path of radius $0.45\,m$ in a magnetic field $B = 1.2\,Wb/m^2$ with a speed of $2.6 \times 10^7\,m/s$. The period of revolution of the $\alpha$-particle is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module