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.
Which one of the following options represents the magnetic field $\vec{B}$ at $O$ due to the current flowing in the given wire segments lying on the $x y$ plane?
A proton is projected with velocity $\overrightarrow{ V }=2 \hat{ i }$ in a region where magnetic field $\overrightarrow{ B }=(\hat{i}+3 \hat{j}+4 \hat{k})\; \mu T$ and electric field $\overrightarrow{ E }=10 \hat{ i } \;\mu V / m .$ Then find out the net acceleration of proton (in $m / s ^{2}$)
One proton beam enters a magnetic field of ${10^{ - 4}}$ $T$ normally, Specific charge = ${10^{11}}\,C/kg.$ velocity = ${10^7}\,m/s$. What is the radius of the circle described by it....$m$
A proton and an alpha particle of the same enter in a uniform magnetic field which is acting perpendicular to their direction of motion. The ratio of the circular paths described by the alpha particle and proton is ....
What is source of magnetic field ?