$A$ charge $Q$ is moving through a distance $d\vec{l}$ in a magnetic field $\vec{B}$. Find the value of the work done by the magnetic field.

An electrically charged particle enters a uniform magnetic field in a direction perpendicular to the field with a velocity $v$. Then, it travels:

An electron,moving in a uniform magnetic field of induction of intensity $\vec{B}$,has its radius directly proportional to

$A$ velocity selector consists of an electric field $\overrightarrow{E} = E \hat{k}$ and a magnetic field $\overrightarrow{B} = B \hat{j}$ with $B = 12 \, mT$. The value of $E$ required for an electron of energy $728 \, eV$ moving along the positive $x$-axis to pass undeflected is: (Given: 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 $28 \text{ cm}$.

An electron of mass $m$ and charge $q$ is travelling with a speed $v$ along a circular path of radius $r$ at right angles to a uniform magnetic field $B$. If the speed of the electron is doubled and the magnetic field is halved,then the resulting path would have a radius of: