$A$ positively charged particle moving due east enters a region of uniform magnetic field directed vertically upwards. The particle will

$A$ positive charge $q$ of mass $m$ is moving along the $+x$ axis with velocity $v$. We wish to apply a uniform magnetic field $B$ for time $\Delta t$ so that the charge reverses its direction,crossing the $y$ axis at a distance $d$. Then:

An electron and a proton enter a magnetic field perpendicularly. Both have the same kinetic energy. Which of the following is true?

When an electron is accelerated by a potential difference of $V$ volts and enters a magnetic field,the force acting on it is $F$. What will be the force acting on the electron when it is accelerated by a potential difference of $5V$ volts and enters the same magnetic field?

$A$ proton is moving with a uniform velocity of $10^{6} \ m/s$ along the $Y$-axis,under the joint action of a magnetic field along the $Z$-axis and an electric field of magnitude $2 \times 10^{4} \ V/m$ along the negative $X$-axis. If the electric field is switched off,the proton starts moving in a circle. The radius of the circle is nearly: (Given: $\frac{e}{m}$ ratio for proton $= 10^{8} \ C/kg$) (in $m$)

$A$ 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 $....... \ m$.