$A$ $10 \; eV$ electron is circulating in a plane at right angles to a uniform magnetic field of magnetic induction $10^{-4} \; Wb/m^2$ $(1.0 \; \text{gauss})$. The orbital radius of the electron is ........ $cm$.

  • A
    $11$
  • B
    $18$
  • C
    $12$
  • D
    $16$

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An electron,moving along the $x-$ axis with an initial energy of $100\, eV$,enters a region of magnetic field $\vec B = (1.5 \times 10^{-3} \, T) \hat k$ at $S$ (See figure). The field extends between $x = 0$ and $x = 2 \, cm$. The electron is detected at the point $Q$ on a screen placed $8 \, cm$ away from the point $S$. The distance $d$ between $P$ and $Q$ (on the screen) is :......$cm$ (electron's charge $= 1.6 \times 10^{-19} \, C$,mass of electron $= 9.1 \times 10^{-31} \, kg$)

$A$ positive,singly ionized atom of mass number $A_M$ is accelerated from rest by a voltage $V = 192 \text{ V}$. Thereafter,it enters a rectangular region of width $w$ with a magnetic field $B_0 = 0.1 \hat{k} \text{ T}$,as shown in the figure. The ion finally hits a detector at a distance $x$ below its starting trajectory.
[Given: Mass of neutron/proton $= (5/3) \times 10^{-27} \text{ kg}$,charge of the electron $= 1.6 \times 10^{-19} \text{ C}$.]
Which of the following option$(s)$ is(are) correct?
$(A)$ The value of $x$ for $H^{+}$ ion is $4 \text{ cm}$.
$(B)$ The value of $x$ for an ion with $A_M = 144$ is $48 \text{ cm}$.
$(C)$ For detecting ions with $1 \leq A_M \leq 196$,the minimum height $(x_1 - x_0)$ of the detector is $52 \text{ cm}$.
$(D)$ The minimum width $w$ of the region of the magnetic field for detecting ions with $A_M = 196$ is $28 \text{ cm}$.

In a chamber,a uniform magnetic field of $6.5 \; G$ $(1 \; G = 10^{-4} \; T)$ is maintained. An electron is shot into the field with a speed of $4.8 \times 10^{6} \; m s^{-1}$ normal to the field. The radius of the circular orbit of the electron is $4.2 \; cm$. Obtain the frequency of revolution of the electron in its circular orbit. Does the answer depend on the speed of the electron? Explain. $(e = 1.6 \times 10^{-19} \; C, m_{e} = 9.1 \times 10^{-31} \; kg)$

Two charged particles traverse identical helical paths in a completely opposite sense in a uniform magnetic field $B = B_0 \hat{k}$.

$A$ particle of charge $q$,mass $m$ enters a region of magnetic field $B$ with velocity $v_0 \widehat{i}$. Find the value of $d$ if the particle emerges from the region of magnetic field at an angle $30^{\circ}$ to its initial velocity:-

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