An ion beam of specific charge $5 \times 10^7$ $coulomb/kg$ enter a uniform magnetic field of $4 \times 10^{-2}\, tesla$ with a velocity $2 \times 10^5\, m/s$ perpendicularly. The radius of the circular path of ions in meter will be

  • A

    $0.10$

  • B

    $0.16$

  • C

    $0.20$

  • D

    $0.25$

Similar Questions

When a charged particle moving with velocity $\vec v$ is subjected to a magnetic field of induction $\vec B$, the force on it is non-zero. This implies that

  • [AIPMT 2006]

Two parallel wires in the plane of the paper are distance $X _0$ apart. A point charge is moving with speed $u$ between the wires in the same plane at a distance $X_1$ from one of the wires. When the wires carry current of magnitude $I$ in the same direction, the radius of curvature of the path of the point charge is $R_1$. In contrast, if the currents $I$ in the two wires have direction opposite to each other, the radius of curvature of the path is $R_2$.

If $\frac{x_0}{x_1}=3$, the value of $\frac{R_1}{R_2}$ is.

  • [IIT 2014]

An electron, a proton, a deuteron and an alpha particle, each having the same speed are in a region of constant magnetic field perpendicular to the direction of the velocities of the particles. The radius of the circular orbits of these particles are respectively $R_e, R_p, R_d \,$ and $\, R_\alpha$. It follows that

A proton and a deutron both having the same kinetic energy, enter perpendicularly into a uniform magnetic field $B$. For motion of proton and deutron on circular path of radius ${R_p}$ and ${R_d}$ respectively, the correct statement is

A particle of charge $q$, mass $m$ enters in a region of magnetic field $B$ with velocity $V_0 \widehat i$. Find the value of $d$ if the particle emerges from the region of magnetic field at an angle $30^o$ to its ititial velocity:-