Question

A 10 eV electron is circulating in a plane at right angles to a uniform field of magnetic induction${\rm{1}}{{\rm{0}}^{{\rm{ - 4}}}}\;{\rm{WB/}}{{\rm{m}}^{\rm{2}}}\left( { = \;1.0\;{\rm{gauss}}} \right)$, the orbital radius of electron is
(a) 11 cm
(b) 18 cm
(c) 12 cm
(d) 16 cm

Answer 1

Hint:To determine the electron's orbital radius, we will equate the centripetal force experienced by the electron due to its circular motion with force experienced by the magnetic field. After this, we will obtain the electron's momentum and put in the expression of centripetal force, so that we will get the orbital radius.

Complete step by step answer:
It is given in the question that energy of the electron in the circular motion is$10\;{\rm{eV}}$ and magnetic induction of the magnetic field is ${10^{ - 4}}\;{\rm{Wb/}}{{\rm{m}}^2}$.
First, we will write the expression of the force due to the perpendicular magnetic field.
${F_{mag}} = Bqv$.... (1)
Here, $ B $ is the magnetic induction, $v$ is the electron's speed, and $q$ is the charge of the electron.
Now we will write the expression of the centripetal force experienced by the electron in the circular motion.
${F_c} = \dfrac{{m{v^2}}}{R}$.... (2)
Here, $R$ is the orbital radius, and $m$ is the mass of the electron.
Equate the equation (1) and (2) to obtain the expression of the orbital radius of the electron.
Therefore, we get
$\begin{array}{l}
\dfrac{{m{v^2}}}{R} = Bqv\\
R = \dfrac{{mv}}{{Bq}}
\end{array}$
We know that the expression of the electron moment is $mv = \sqrt {2mkE} $, so we will use this in the above expression, so the above expression becomes,
$R = \dfrac{{\sqrt {2mKE} }}{{Bq}}$
Here $KE$ is the energy of the electron.
We know that the mass of the electron is $m = 9.1 \times {10^{ - 31}}\;{\rm{kg}}$, charge of the electron is $q = 1.6 \times {10^{ - 19}}\;{\rm{C}}$ so will substitute these values and given values in the above expression, so that we can obtain orbital radius.
Therefore, we get
\[\begin{array}{l}
R = \dfrac{{\sqrt {2\left( {9.1 \times {{10}^{ - 31}}\;{\rm{kg}}} \right)\left( {10\;{\rm{eV}} \times \dfrac{{1.6 \times {{10}^{ - 19}}\;{\rm{J}}}}{{1\;{\rm{eV}}}}} \right)} }}{{\left( {{{10}^{ - 4}}\;{\rm{Wb/}}{{\rm{m}}^{\rm{2}}}} \right)\left( {1.6 \times {{10}^{ - 19}}\;{\rm{C}}} \right)}}\\
R = 0.1066\;{\rm{m}} \times \dfrac{{100\;{\rm{cm}}}}{{1\;{\rm{m}}}}\\
{\rm{R = 10}}{\rm{.66}}\;{\rm{cm}} \approx 11\;{\rm{cm}}
\end{array}\]
Therefore, the orbital radius of the electron is $11\;{\rm{cm}}$ and option (1) is correct.

Note: In this problem, when the electron is circulating in a uniform magnetic field plane, then the centripetal force experienced by the electron will be equal to the force exerted by the magnetic force because the magnetic force acts as centripetal force.