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Question: An electron is revolving around a proton in a circular orbit of diameter \(1\overset{{}^\circ }{\mat...

An electron is revolving around a proton in a circular orbit of diameter 1A∘ 1\overset{{}^\circ }{\mathop{\text{A}}}\,. If it produces a magnetic field of 14 Wb/m214\ Wb/{{m}^{2}} at the proton, then its angular velocity will be about
A. 8.75×1016 rad/s8.75\times {{10}^{16}}\ rad/s
B. 1010  rad/s{{10}^{10\ }}\ rad/s
C. 4×1015 rad/s4\times {{10}^{15}}\ rad/s
D. 1015  rad/s{{10}^{15\ }}\ rad/s

Explanation

Solution

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

Formula used: i=qti=\dfrac{q}{t}, t=2πwt=\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 1A∘ 1\overset{{}^\circ }{\mathop{\text{A}}}\,. If it produces a magnetic field of 14 Wb/m214\ 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=qti=\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=2πwt=\dfrac{2\pi }{w} ………..(ii)
Where t is time and ww is angular velocity.
Now, in question it is given that, the magnetic field produced due to the revolution of electron is 14 Wb/m214\ 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=μ0i2RB=\dfrac{{{\mu }_{0}}i}{2R} ………………….(iii)
Where, i is current, B is magnetic field and R is radius, μ0=4π×10−7{{\mu }_{0}}=4\pi \times {{10}^{-7}} .
Now, the value of magnetic field is 14 Wb/m214\ Wb/{{m}^{2}} and diameter is 1A∘ 1\overset{{}^\circ }{\mathop{\text{A}}}\,, substituting this values in the expression we will get,
B=μ0q2RtB=\dfrac{{{\mu }_{0}}q}{2Rt} from expression (i)
⇒B=μ0qw2R2π\Rightarrow B=\dfrac{{{\mu }_{0}}qw}{2R2\pi } from expression (ii)
Now, substituting all the given values we will get,
⇒14=4π×10−7qw2D22π⇒14=4π×10−7qw2×10−10×2π\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=D2R=\dfrac{D}{2} and 1A∘ =10−101\overset{{}^\circ }{\mathop{\text{A}}}\,={{10}^{-10}} , now, q=e−=1.6×10−19q={{e}^{-}}=1.6\times {{10}^{-19}} , so, now the expression will be,
⇒14=10−7×1.6×10−19×w10−10⇒w=14×10−1010−7×1.6×10−19\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}}}
⇒w=14×10−1010−7×1.6×10−19=1416×1017=0.875×1017\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}}
⇒w=8.75×1016 rad/s\Rightarrow w=8.75\times {{10}^{16}}\ rad/s
Hence, angular velocity is 8.75×1016 rad/s8.75\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=μ0nitB={{\mu }_{0}}nit instead of B=μ0i2RB=\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.