An electron is moving around a proton in an orbit of radius $1 \overset{˚}{A}$ and produces $16 W b / m^{2}$ of magnetic field at the centre, then find the angular velocity of electron:

Question

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Answer 1

$B = 16 W b / m^{2}$
$R = \overset{˚}{A}$
as $B = \frac{μ _{0} I}{4 π R ^{2}} \oint d L B = \frac{μ _{0} I}{4 π R ^{2}} \times 2 π R = \frac{μ _{0} I}{2 R}$
as $I = \frac{e}{t}$ or $I = \frac{e \times ω}{π}$ (as $\frac{2 π}{ω} = t$ )
So, $16 = \frac{4 π \times 1 0 ^{- 7} \times ( 1 . 6 \times 1 0 ^{19} ) \times \frac{ω}{2 π}}{2 \times 1 0 ^{- 10}} \Rightarrow ω = \frac{1 6 \times 1 0 ^{- 10}}{1 . 6 \times 1 0 ^{- 26}} \Rightarrow ω = 10^{17}$ rad/sec

Answer 2

B = 16 W b / m^{2}

R = \overset{˚}{A}

as

B = \frac{μ _{0} I}{4 π R ^{2}} \oint d L B = \frac{μ _{0} I}{4 π R ^{2}} \times 2 π R = \frac{μ _{0} I}{2 R}

as

I = \frac{e}{t}

or

I = \frac{e \times ω}{π}

(as

\frac{2 π}{ω} = t

)

So,

16 = \frac{4 π \times 1 0 ^{- 7} \times ( 1 . 6 \times 1 0 ^{19} ) \times \frac{ω}{2 π}}{2 \times 1 0 ^{- 10}} \Rightarrow ω = \frac{1 6 \times 1 0 ^{- 10}}{1 . 6 \times 1 0 ^{- 26}} \Rightarrow ω = 10^{17}

rad/sec