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If a charged particle enters a uniform magnetic field normally with a certain velocity,then the time period of revolution of the particle

An electron experiences a force $\vec{F} = (4.0\,\hat{i} + 3.0\,\hat{j}) \times 10^{-13}\,N$ in a uniform magnetic field when its velocity is $\vec{v}_1 = 2.5\,\hat{k} \times 10^7\,m/s$. When the velocity is redirected to $\vec{v}_2 = (1.5\,\hat{i} - 2.0\,\hat{j}) \times 10^7\,m/s$,the magnetic force on the electron is zero. The magnetic field $\vec{B}$ is:

$A$ particle having the same charge as an electron moves in a circular path of radius $0.5 \, cm$ under the influence of a magnetic field of $0.5 \, T$. If an electric field of $100 \, V/m$ makes it move in a straight path,then the mass of the particle is (given charge of electron $= 1.6 \times 10^{-19} \, C$):

$A$ proton (mass $m$,charge $q$) accelerated by a potential difference $V$ flies through a uniform transverse magnetic field $B$. The field occupies a region of space of width $d$. If $\alpha$ is the angle of deviation of the proton from the initial direction of motion (see figure),the value of $\sin \alpha$ will be:

$A$ proton and a deuteron are both accelerated through the same potential difference and enter a magnetic field perpendicular to the direction of the field. If the deuteron follows a path of radius $R$,assuming the neutron and proton masses are nearly equal,the radius of the proton's path will be

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