If the direction of the initial velocity of the charged particle is neither along nor perpendicular to that of the magnetic field,then the orbit will be

$A$ stream of electrons is projected horizontally to the right. $A$ straight conductor carrying a current is supported parallel to the electron stream and above it. If the current in the conductor is from left to right,then what will be the effect on the electron stream?

$A$ 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 $= 0.96 \times 10^8 \text{ C kg}^{-1}$)

In an experiment,electrons are accelerated from rest by applying a voltage of $500 \, V$. Calculate the radius of the path if a magnetic field of $100 \, mT$ is then applied. [Charge of the electron $= 1.6 \times 10^{-19} \, C$,Mass of the electron $= 9.1 \times 10^{-31} \, kg$]

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

$A$ charged particle is moving in a uniform magnetic field. It penetrates a layer of lead and thereby loses half of its kinetic energy. What happens to the radius of curvature of its path?