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 :

  • A

    $4 \,km$

  • B

    $4 \,m$

  • C

    $4 \,cm$

  • D

    $4 \,mm$

Similar Questions

Two identical charged particles enter a uniform magnetic field with same speed but at angles $30^o$ and $60^o$ with field. Let $a, b$ and $c$ be the ratio of their time periods, radii and pitches of the helical paths than

A particle of mass $m$ and charge $\mathrm{q}$, moving with velocity $\mathrm{V}$ enters Region $II$ normal to the boundary as shown in the figure. Region $II$ has a uniform magnetic field B perpendicular to the plane of the paper. The length of the Region $II$ is $\ell$. Choose the correct choice$(s)$.

Figure: $Image$

$(A)$ The particle enters Region $III$ only if its velocity $V>\frac{q / B}{m}$

$(B)$ The particle enters Region $III$ only if its velocity $\mathrm{V}<\frac{\mathrm{q} / \mathrm{B}}{\mathrm{m}}$

$(C)$ Path length of the particle in Region $II$ is maximum when velocity $V=\frac{q / B}{m}$

$(D)$ Time spent in Region $II$ is same for any velocity $V$ as long as the particle returns to Region $I$

  • [IIT 2008]

Mark the correct statement

In a chamber, a uniform magnetic field of $6.5 \;G \left(1 \;G =10^{-4} \;T \right)$ is maintained. An electron is shot into the field with a speed of $4.8 \times 10^{6} \;m s ^{-1}$ normal to the field.the radius of the circular orbit of the electron is $4.2 \;cm$. obtain the frequency of revolution of the electron in its circular orbit. Does the answer depend on the speed of the electron? Explain.

$\left(e=1.5 \times 10^{-19} \;C , m_{e}=9.1 \times 10^{-31}\; kg \right)$

A charged particle of charge $q$ and mass $m$, gets deflected through an angle $\theta$ upon passing through a square region of side $a$, which contains a uniform magnetic field $B$ normal to its plane. Assuming that the particle entered the square at right angles to one side, what is the speed of the particle?

  • [KVPY 2010]