An electron (mass $= 9 \times 10^{-31}\,kg$. Charge $= 1.6 \times 10^{-19}\,C$) whose kinetic energy is $7.2 \times 10^{-18}$ $joule$ is moving in a circular orbit in a magnetic field of $9 \times 10^{-5} \,weber/m^2$. The radius of the orbit is.....$cm$
$1.25$
$2.5$
$12.5 $
$25$
A charge particle of $2\,\mu\,C$ accelerated by a potential difference of $100\,V$ enters a region of uniform magnetic field of magnitude $4\,m\,T$ at right angle to the direction of field. The charge particle completes semicircle of radius $3\,cm$ inside magnetic field. The mass of the charge particle is $........\times 10^{-18}\,kg$.
An electron with energy $880 \,eV$ enters a uniform magnetic field of induction $2.5 \times 10^{-3}\,T$. The radius of path of the circle will approximately be :
Proton with kinetic energy of $1\;MeV$ moves from south to north. It gets an acceleration of $10^{12}\; \mathrm{m} / \mathrm{s}^{2}$ by an applied magnetic field (west to east). The value of magnetic field :.......$mT$ (Rest mass of proton is $1.6 \times 10^{-27} \;\mathrm{kg}$ )
If a proton, deutron and $\alpha - $ particle on being accelerated by the same potential difference enters perpendicular to the magnetic field, then the ratio of their kinetic energies is
A proton is projected with velocity $\overrightarrow{ V }=2 \hat{ i }$ in a region where magnetic field $\overrightarrow{ B }=(\hat{i}+3 \hat{j}+4 \hat{k})\; \mu T$ and electric field $\overrightarrow{ E }=10 \hat{ i } \;\mu V / m .$ Then find out the net acceleration of proton (in $m / s ^{2}$)