(A) The radius of a charged particle moving in a magnetic field is given by $r = \frac{mv}{qB}$.Since kinetic energy $K = \frac{1}{2}mv^2$,we have $mv

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(A) The radius of a charged particle moving in a magnetic field is given by $r = \frac{mv}{qB}$.Since kinetic energy $K = \frac{1}{2}mv^2$,we have $mv = \sqrt{2mK}$.Thus,$r = \frac{\sqrt{2mK}}{qB}$.For a deuteron $(d)$ and a proton $(p)$:Mass of deuteron $m_d = 2m_p$,charge $q_d = e$.Mass of proton $m_p = m_p$,charge $q_p = e$.Given kinetic energies are equal $(K_d = K_p = K)$.Ratio of radii: $\frac{r_d}{r_p} = \frac{\sqrt{2m_d K} / eB}{\sqrt{2m_p K} / eB} = \sqrt{\frac{m_d}{m_p}} = \sqrt{\frac{2m_p}{m_p}} = \sqrt{2}$.Comparing $\frac{r_d}{r_p} = \sqrt{2} : 1$ with $\sqrt{x} : 1$,we get $x = 2$.

Class 12 Physics Moving Charges and Magnetism Motion of Charged Particle In Magnetic Field

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$A$ deuteron and a proton moving with equal kinetic energy enter into a uniform magnetic field at a 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 $\sqrt{x} : 1$ where $x$ is ..........

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Motion of Charged Particle In Magnetic Field

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$A$ deuteron and a proton moving with equal kinetic energy enter into a uniform magnetic field at a 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 $\sqrt{x} : 1$ where $x$ is ..........

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$A$ proton accelerated by a potential difference of $500 \; kV$ moves through a transverse magnetic field of $0.51 \; T$ as shown in the figure. The angle $\theta$ through which the proton deviates from the initial direction of its motion is......$^o$

$A$ proton is moving perpendicular to a uniform magnetic field of $2.5 \ T$ with $2 \ MeV$ kinetic energy. The force on the proton is . . . . . . $N$. (Mass of proton $= 1.6 \times 10^{-27} \ kg$,charge of proton $= 1.6 \times 10^{-19} \ C$)

$A$ proton of mass $m$ moving with a speed $v$ ($v << c$, where $c$ is the velocity of light in vacuum) completes a circular orbit in time $T$ in a uniform magnetic field. If the speed of the proton is increased to $\sqrt{2}v$, what will be the time needed to complete the circular orbit?

$A$ proton,a deuteron and an $\alpha$-particle having the same kinetic energy are moving in circular trajectories in a constant magnetic field. If ${r_p}$,${r_d}$ and ${r_\alpha}$ denote respectively the radii of the trajectories of these particles,then:

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