$A$ charged particle moves through a magnetic field in a direction perpendicular to it. Then the

$A$ 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 $rad$)

Under the influence of a uniform magnetic field,a charged particle is moving in a circle of radius $R$ with constant speed $v$. The time period of the motion

$A$ positively charged particle enters a region of a uniform transverse magnetic field as shown in the figure. Find the net deviation in the path of the particle.

At $t = 0$,a charge $q$ is at the origin and moving in the $y$-direction with velocity $\vec{v} = v\hat{j}$. The charge moves in a magnetic field that is for $y > 0$ out of the page and given by $B_1\hat{k}$ and for $y < 0$ into the page and given by $-B_2\hat{k}$. The charge's subsequent trajectory is shown in the sketch. From this information,we can deduce that:

An electron enters a magnetic field whose direction is perpendicular to the velocity of the electron. Then