$A$ 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.

$A$ proton accelerated by a potential difference of $500 \ kV$ flies through a uniform transverse magnetic field of $0.1 \ T$. The field is spread over a region of $1.0 \ cm$ thickness. The angle through which the proton gets deviated from its original direction is (Proton mass $= 1.6 \times 10^{-27} \ kg$ and charge of proton $= 1.6 \times 10^{-19} \ C$) (in $rad$)

$A$ uniform electric field and a uniform magnetic field are acting along the same direction in a certain region. If an electron is projected in the region such that its velocity is pointed along the direction of fields,then the electron

$A$ uniform magnetic field acts at right angles to the direction of motion of electrons. As a result,the electron moves in a circular path of radius $2\, cm$. If the speed of the electrons is doubled,then the radius of the circular path will be.....$cm$.

$A$ strong magnetic field is applied on a stationary electron,then

In the $xy$-plane,the region $y > 0$ has a uniform magnetic field $B_1 \hat{k}$ and the region $y < 0$ has another uniform magnetic field $B_2 \hat{k}$. $A$ positively charged particle is projected from the origin along the positive $y$-axis with speed $v_0 = \pi \text{ m s}^{-1}$ at $t = 0$,as shown in the figure. Neglect gravity in this problem. Let $t = T$ be the time when the particle crosses the $x$-axis from below for the first time. If $B_2 = 4 B_1$,the average speed of the particle,in $\text{m s}^{-1}$,along the $x$-axis in the time interval $T$ is. . . . . .