Two particles $X$ and $Y$ having equal charges,are accelerated through the same potential difference. They enter a region of a uniform magnetic field and describe a circular path of radii $r_1$ and $r_2$ respectively. The ratio of the mass of $X$ to that of $Y$ is

$A$ proton is projected with a uniform velocity $v$ along the axis of a current-carrying solenoid,then

$A$ proton is projected with a velocity $10^7\, m/s$,at right angles to a uniform magnetic field of induction $100\, mT$. The time (in seconds) taken by the proton to traverse a $90^o$ arc is (given $m_p = 1.65 \times 10^{-27}\, kg$ and $q_p = 1.6 \times 10^{-19}\, C$).

An $\alpha$-particle (mass $4 \text{ amu}$) and a singly charged sulfur ion (mass $32 \text{ amu}$) are initially at rest. They are accelerated through a potential $V$ and then allowed to pass into a region of uniform magnetic field which is normal to the velocities of the particles. Within this region,the $\alpha$-particle and the sulfur ion move in circular orbits of radii $r_\alpha$ and $r_s$,respectively. The ratio $(r_s / r_\alpha)$ is:

$A$ particle of mass $m = 1.67 \times 10^{-27} \, kg$ and charge $q = 1.6 \times 10^{-19} \, C$ enters a region of uniform magnetic field of strength $B = 1 \, T$ along the direction shown in the figure. The particle leaves the magnetic field at point $D$. Calculate the distance $CD$. (Assume the velocity of the particle $v = 10^7 \, m/s$)

$A$ very long straight wire carries a current $I$. At the instant when a charge $+Q$ at point $P$ has velocity $\vec{V}$,as shown,the force on the charge is