An electron falling freely under the influence of gravity enters a uniform magnetic field directed towards the south. The electron is initially deflected towards:

  • A
    east
  • B
    west
  • C
    north
  • D
    south

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$A$ proton,a deuteron and an $\alpha$-particle having the same kinetic energy are moving in circular trajectories in a constant magnetic field. If ${r_p}$,${r_d}$ and ${r_\alpha}$ denote respectively the radii of the trajectories of these particles,then:

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$A$ particle of mass $M$ and positive charge $Q$,moving with a constant velocity $\vec{u}_1 = 4\hat{i} \text{ m/s}$,enters a region of uniform static magnetic field normal to the $x-y$ plane. The region of the magnetic field extends from $x = 0$ to $x = L$ for all values of $y$. After passing through this region,the particle emerges on the other side after $10 \text{ ms}$ with a velocity $\vec{u}_2 = 2(\sqrt{3}\hat{i} + \hat{j}) \text{ m/s}$. The correct statement$(s)$ is (are):
$(A)$ The direction of the magnetic field is $-z$ direction.
$(B)$ The direction of the magnetic field is $+z$ direction.
$(C)$ The magnitude of the magnetic field is $\frac{50\pi M}{3Q}$ units.
$(D)$ The magnitude of the magnetic field is $\frac{100\pi M}{3Q}$ units.

$A$ 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 $= 1.6 \times 10^{-19} \, C$):

$A$ beam of protons moving with a velocity $1.6 \times 10^5 \ m/s$ enters a uniform magnetic field of $\frac{\pi}{10} \ T$ at an angle $60^{\circ}$ to the direction of the field. The pitch of the helical path of the protons is (mass of proton $= 1.6 \times 10^{-27} \ kg$)

$A$ proton, a deuteron (nucleus of ${ }_1 H^2$) and an $\alpha$-particle with the same kinetic energy enter a region of uniform magnetic field moving at right angles to the field. The ratio of the radii of their circular paths is:

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