If the magnetic field is parallel to the positive $y-$axis and the charged particle is moving along the positive $x-$axis (Figure), which way would the Lorentz force be for
$(a)$ an electron (negative charge),
$(b)$ a proton (positive charge).
The velocity $v$ of particle is along the $x$ -axis, while $B ,$ the magnetic fleld is along the $y$ -axis, so $v \times B$ is along the $z$ -axis (screw rule or right-hand thumb rule). So, $(a)$ for electron it will be along $-z$ axis. $(b)$ for a positive charge (proton) the force is along $+z$ axis.
A proton is projected with a velocity $10^7\, m/s$, at right angles to a uniform magnetic field of induction $100\, mT$. The time (in second) taken by the proton to traverse $90^o$ arc is $(m_p = 1.65\times10^{-27}\, kg$ and $q_p = 1.6\times10^{-19}\, C)$
A uniform electric field and a uniform magnetic field are produced, pointed in the same direction. An electron is projected with its velocity pointing in the same direction
An electron and a proton are moving on straight parallel paths with same velocity. They enter a semi-infinite region of uniform magnetic field perpendicular to the velocity. Which of the following statement$(s)$ is/are true?
$(A)$ They will never come out of the magnetic field region.
$(B)$ They will come out travelling along parallel paths.
$(C)$ They will come out at the same time.
$(D)$ They will come out at different times.
The magnetic field is uniform for $y>0$ and points into the plane. The magnetic field is uniform and points out of the plane for $y<0$. A proton denoted by filled circle leaves $y=0$ in the $-y$-direction with some speed as shown below.Which of the following best denotes the trajectory of the proton?
A particle of mass $'m'$ and carrying a charge $'q'$ enters with a velocity $'v'$ perpendicular to a uniform magnetic field. The time period of rotation of the particle