$A$ proton is projected with a velocity $10^7\, m/s$,at right angles to a uniform magnetic field of induction $100\, mT$. The time (in seconds) taken by the proton to traverse a $90^o$ arc is (given $m_p = 1.65 \times 10^{-27}\, kg$ and $q_p = 1.6 \times 10^{-19}\, C$).

$A$ rectangular region of dimensions $(\omega \times l)$ where $\omega \ll l$ has a constant magnetic field $B$ directed into the plane of the paper as shown in the figure. On one side,the region is bounded by a screen. On the other side,positive ions of mass $m$ and charge $q$ are accelerated from rest by a parallel plate capacitor at a constant potential difference $V$ and enter the magnetic field region through a small hole. Which one of the following statements is correct regarding the charge $q$ on the ions that hit the screen?

$A$ particle of charge $q = -16 \times 10^{-18} \, C$ moving with velocity $v = 10 \, m/s$ along the $x$-axis enters a region where a magnetic field of induction $B$ is along the $y$-axis,and an electric field of magnitude $E = 10^4 \, V/m$ is along the negative $z$-axis. If the charged particle continues moving along the $x$-axis,the magnitude of $B$ is:

When an electron placed in a uniform magnetic field is accelerated from rest through a potential difference $V_1$,it experiences a force $F$. If the potential difference is changed to $V_2$,the force experienced by the electron in the same magnetic field is $2F$. Then,the ratio of potential differences $\frac{V_2}{V_1}$ is:

$A$ proton and an $\alpha$-particle are simultaneously projected in opposite directions into a region of uniform magnetic field of $2 \text{ mT}$ perpendicular to the direction of the field. After some time,it is found that the velocity of the proton has changed in direction by $90^{\circ}$. Then,at this time,the angle between the velocity vectors of the proton and the $\alpha$-particle is (in $^{\circ}$)

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