$A$ beam of ions with velocity $2 \times 10^5 \ m/s$ enters normally into a uniform magnetic field of $4 \times 10^{-2} \ T$. If the specific charge of the ion is $5 \times 10^7 \ C/kg$,then the radius of the circular path described will be $....... \ m$.

The charge on a particle $Y$ is double the charge on particle $X$. These two particles $X$ and $Y$,after being accelerated through the same potential difference,enter a region of uniform magnetic field and describe circular paths of radii $R_1$ and $R_2$ respectively. The ratio of the mass of $X$ to that of $Y$ is

$A$ proton of velocity $\vec{v} = (3 \hat{i} + 2 \hat{j}) \text{ ms}^{-1}$ enters a magnetic field of induction $\vec{B} = (2 \hat{j} + 3 \hat{k}) \text{ T}$. The acceleration produced in the proton in $\text{ms}^{-2}$ is (Specific charge of proton $= 0.96 \times 10^8 \text{ C kg}^{-1}$)

$A$ charged particle, when entering a uniform magnetic field, moves in a helical path. If its angular velocity is $4 \pi \times 10^6 \text{ rad s}^{-1}$ and its velocity in the direction of the magnetic field is $3 \times 10^5 \text{ m s}^{-1}$, then the pitch of the helix is: (in $\text{ cm}$)

$A$ 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 moving with a velocity of $2.5 \times 10^7 \, m/s$ enters a magnetic field of intensity $2.5 \, T$ making an angle of $30^o$ with the magnetic field. The force on the proton is: