$A$ particle of mass $m = 1.67 \times 10^{-27} \, kg$ and charge $q = 1.6 \times 10^{-19} \, C$ enters a region of uniform magnetic field of strength $B = 1 \, T$ along the direction shown in the figure. The particle leaves the magnetic field at point $D$. Calculate the distance $CD$. (Assume the velocity of the particle $v = 10^7 \, m/s$)

$A$ proton, a deuteron, and an $\alpha$-particle with the same kinetic energy enter a uniform magnetic field at a right angle to the magnetic field. The ratio of the radii of their respective circular paths is

Consider a long thin conducting wire carrying a uniform current $I$. $A$ particle having mass $M$ and charge $q$ is released at a distance $a$ from the wire with a speed $v_0$ along the direction of current in the wire. The particle gets attracted to the wire due to magnetic force. The particle turns round when it is at distance $x$ from the wire. The value of $x$ is [$\mu_0$ is vacuum permeability]

An electron accelerated through a potential difference $V$ passes through a uniform transverse magnetic field and experiences a force $F$. If the accelerating potential is increased to $2V$,the electron in the same magnetic field will experience a force:

An infinitely long straight conductor carries a current of $5 \, A$ as shown. An electron is moving with a speed of $10^{5} \, m/s$ parallel to the conductor. The perpendicular distance between the electron and the conductor is $20 \, cm$ at an instant. Calculate the magnitude of the force experienced by the electron at that instant in $\times 10^{-20} \, N$.

$A$ horizontal conductor is oriented north-south and carries some current. $A$ positively charged particle located vertically above it and having a velocity directed northward experiences an upward force. What is the direction of the force if this charged particle were located to the east of the conductor and had a velocity directed towards the conductor?