$A$ proton (mass $m = 1.67 \times 10^{-27} \, kg$ and charge $q = 1.6 \times 10^{-19} \, C$) enters perpendicular to a magnetic field of intensity $B = 2 \, Wb/m^2$ with a velocity $v = 3.4 \times 10^7 \, m/s$. The acceleration of the proton is:

An electron,a proton,a deuteron,and an $\alpha -$ particle enter a uniform magnetic field normally with the same velocity. Which particle will have the maximum rotational frequency?

An electron is projected along the axis of a circular conductor carrying current $I$. The electron will experience:

$A$ charged particle of mass $m$ and charge $q$ describes circular motion of radius $r$ in a uniform magnetic field of strength $B$. The frequency of revolution is

If a proton of kinetic energy $8.35 \text{ MeV}$ enters a uniform magnetic field of $10 \text{ T}$ at right angles to the direction of the field,then the force acting on the proton is (Mass of proton $= 1.67 \times 10^{-27} \text{ kg}$ and charge of proton $= 1.6 \times 10^{-19} \text{ C}$)

At $t = 0$,a charge $q$ is at the origin and moving in the $y$-direction with velocity $\vec{v} = v\hat{j}$. The charge moves in a magnetic field that is for $y > 0$ out of the page and given by $B_1\hat{k}$ and for $y < 0$ into the page and given by $-B_2\hat{k}$. The charge's subsequent trajectory is shown in the sketch. From this information,we can deduce that: