$A$ charged particle moving in a magnetic field experiences a resultant force:

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

Two ions have equal masses but one is singly ionized and the second is doubly ionized. They are projected from the same place in a uniform transverse magnetic field with the same velocity. Then:
$(a)$ Both ions will move along circles of equal radii.
$(b)$ The radius of the circle described by the singly ionized charge is double the radius of the circle described by the doubly ionized charge.
$(c)$ Both circles do not touch each other.
$(d)$ Both circles touch each other.

In the figure shown,a charge $q$ of mass $m$ moving with a velocity $v$ along the $x$-axis enters a region of uniform magnetic field $B$ directed into the page. If the particle is able to enter the region $x > b$,then the velocity $v$ must be greater than:

An electron is accelerated by a potential difference of $12000\, V$. It then enters a uniform magnetic field of $10^{-3}\, T$ applied perpendicular to the path of the electron. Find the radius of the path. Given: mass of electron $m = 9 \times 10^{-31}\, kg$ and charge on electron $q = 1.6 \times 10^{-19}\, C$.

An electron beam travels with a velocity of $1.6 \times 10^7 \ m/s$ perpendicularly to a magnetic field of intensity $0.1 \ T$. Calculate the radius of the path of the electron beam. (Given: mass of electron $m_e = 9 \times 10^{-31} \ kg$)