Question

An electron is revolving around a proton in a circular orbit of diameter $1\overset{{}^\circ }{\mathop{\text{A}}}\,$. If it produces a magnetic field of $14\ Wb/{{m}^{2}}$ at the proton, then its angular velocity will be about
A. $8.75\times {{10}^{16}}\ rad/s$
B. ${{10}^{10\ }}\ rad/s$
C. $4\times {{10}^{15}}\ rad/s$
D. ${{10}^{15\ }}\ rad/s$

Answer 1

Hint: Current produced is equal to the electron passing per unit time, which can be given mathematically as, $i=\dfrac{q}{t}$ and we also know that relation between time and angular velocity can be given by the formula, $t=\dfrac{2\pi }{w}$ . So, using these formulas we will find the value of angular velocity.

Formula used: $i=\dfrac{q}{t}$, $t=\dfrac{2\pi }{w}$

Complete step by step answer:
In question it is given that an electron is revolving around a proton in a circular orbit of diameter $1\overset{{}^\circ }{\mathop{\text{A}}}\,$. If it produces a magnetic field of $14\ Wb/{{m}^{2}}$ at the proton and we are asked to find the angular velocity, so first of all we will understand the relation between current, electron and time.
Current passing through the wire is equal to the number of electrons passing through the wire in time t, this can be shown mathematically as,
$i=\dfrac{q}{t}$ …………..(i)
Where i is current, q is electric charge and t is time.
Now, we know that the relation between time and angular velocity can be given by,
$t=\dfrac{2\pi }{w}$ ………..(ii)
Where t is time and $w$ is angular velocity.
Now, in question it is given that, the magnetic field produced due to the revolution of electron is $14\ Wb/{{m}^{2}}$, so, we can say that the electron forms a circular loop around proton while revolving and we know that magnetic field produced due to circular loop can be given by the formula,
$B=\dfrac{{{\mu }_{0}}i}{2R}$ ………………….(iii)
Where, i is current, B is magnetic field and R is radius, ${{\mu }_{0}}=4\pi \times {{10}^{-7}}$ .
Now, the value of magnetic field is $14\ Wb/{{m}^{2}}$ and diameter is $1\overset{{}^\circ }{\mathop{\text{A}}}\,$, substituting this values in the expression we will get,
$B=\dfrac{{{\mu }_{0}}q}{2Rt}$ from expression (i)
$\Rightarrow B=\dfrac{{{\mu }_{0}}qw}{2R2\pi }$ from expression (ii)
Now, substituting all the given values we will get,
$\Rightarrow 14=\dfrac{4\pi \times {{10}^{-7}}qw}{2\dfrac{D}{2}2\pi }\Rightarrow 14=\dfrac{4\pi \times {{10}^{-7}}qw}{2\times {{10}^{-10}}\times 2\pi }$
As, $R=\dfrac{D}{2}$ and $1\overset{{}^\circ }{\mathop{\text{A}}}\,={{10}^{-10}}$ , now, $q={{e}^{-}}=1.6\times {{10}^{-19}}$ , so, now the expression will be,
\[\Rightarrow 14=\dfrac{{{10}^{-7}}\times 1.6\times {{10}^{-19}}\times w}{{{10}^{-10}}}\Rightarrow w=\dfrac{14\times {{10}^{-10}}}{{{10}^{-7}}\times 1.6\times {{10}^{-19}}}\]
\[\Rightarrow w=\dfrac{14\times {{10}^{-10}}}{{{10}^{-7}}\times 1.6\times {{10}^{-19}}}=\dfrac{14}{16}\times {{10}^{17}}=0.875\times {{10}^{17}}\]
\[\Rightarrow w=8.75\times {{10}^{16}}\ rad/s\]
Hence, angular velocity is \[8.75\times {{10}^{16}}\ rad/s\].
Thus, option (a) is correct.

Note: Student might not consider the loop formed due to revolution of electron around proton and they might use the formula $B={{\mu }_{0}}nit$ instead of $B=\dfrac{{{\mu }_{0}}i}{2R}$ . So, due to that they won’t be able to find the answer and sum may go wrong, so students must be careful while solving such problems.