An electron with energy $880 \,eV$ enters a uniform magnetic field of induction $2.5 \times 10^{-3} \,T$. The radius of the circular path will approximately be:

An electron beam travels with a velocity of $1.6 \times 10^7 \ m/s$ perpendicularly to a magnetic field of intensity $0.1 \ T$. Calculate the radius of the path of the electron beam. (Given: mass of electron $m_e = 9 \times 10^{-31} \ kg$)

$A$ particle with charge $q$,moving with a momentum $p$,enters a uniform magnetic field normally. The magnetic field has magnitude $B$ and is confined to a region of width $d$,where $d < \frac{p}{Bq}$. The particle is deflected by an angle $\theta$ in crossing the field.

$A$ particle of mass $m = 1.67 \times 10^{-27} \, kg$ and charge $q = 1.6 \times 10^{-19} \, C$ enters a region of uniform magnetic field of strength $B = 1 \, T$ as shown in the figure. The angle of incidence is $45^{\circ}$. The radius of the circular portion of the path is:

$A$ charged particle is moving in a circular orbit of radius $6 \, cm$ with a uniform speed of $3 \times 10^6 \, m/s$ under the action of a uniform magnetic field $2 \times 10^{-4} \, Wb/m^2$ which is at right angles to the plane of the orbit. The charge to mass ratio of the particle is

$A$ particle of charge $1.0 \times 10^{-16} \ C$ moves through a uniform magnetic field $B=B_0(\hat{i}+4 \hat{j}) \ T$. The particle velocity at some instant is $v=(2 \hat{i}+4 \hat{j}) \ ms^{-1}$ and the magnetic force acting on it is $3 \times 10^{-16} \hat{k} \ N$. The magnitude of $B_0$ is (in $T$)