$A$ proton beam enters a magnetic field of $10^{-4} \ T$ normally. Given the specific charge $\frac{q}{m} = 10^{11} \ C/kg$ and velocity $v = 10^7 \ m/s$,what is the radius of the circular path described by the beam in meters?

$A$ particle of mass $m$ and charge $q$,accelerated by a potential difference $V$,enters a region of a uniform transverse magnetic field $B$. If $d$ is the thickness of the region of $B$,the angle $\theta$ through which the particle deviates from the initial direction on leaving the region is given by

$A$ charged particle of $2\,\mu\,C$ accelerated by a potential difference of $100\,V$ enters a region of uniform magnetic field of magnitude $4\,mT$ at a right angle to the direction of the field. The charged particle completes a semicircle of radius $3\,cm$ inside the magnetic field. The mass of the charged particle is $........\times 10^{-18}\,kg$.

An electron is moving with a speed of $10^8 \ m/s$ perpendicular to a uniform magnetic field of intensity $B$. Suddenly,the intensity of the magnetic field is reduced to $B/2$. The radius of the path becomes from the original value of $r$:

$A$ particle of mass $0.6 \,g$ and having charge of $25 \,nC$ is moving horizontally with a uniform velocity $1.2 \times 10^4 \,ms^{-1}$ in a uniform magnetic field. If the particle moves in a straight line, the value of the magnetic induction is $(g=10 \,ms^{-2})$.

$A$ proton, an electron, and an $\alpha$-particle enter at right angles to a uniform magnetic field with the same velocity. If $R_p$, $R_e$, and $R_{\alpha}$ are the radii of the circular paths of these particles, then: