An electron moves around the nucleus in a hydrogen atom of radius 0.51 Å, with a velocity of 2 x 105 m/s. Calculate the following:
(i) the equivalent current due to orbital motion of electron,
(ii) the magnetic field produced at the centre of the nucleus
(iii) the magnetic moment associated with the electron.

Solution

Given,
Radius, r = 0.51 Å = 0.51 x 10^–10 m
velocity, v = 2 x 10⁵ m/s

(i) Equivalent current due to orital motion of electron is given by,

$I = \frac{e}{t} = \frac{e}{2 πr / v} = \frac{ev}{2 πr}$

$I = \frac{1.6 \times 10^{- 19} \times 2 \times 10^{5}}{2 \times 3.14 \times 0.51 \times 10^{- 10}}$

$I = \frac{3.2 \times 10^{- 4}}{3.2028} = 0.99 \times 10^{- 4} = 10^{- 4} A$

(ii) Magnetic field produced at the centre of the electron is,
$B = μnI$

$B = 4 π \times 10^{- 7} \times 1 \times 10^{- 4}$

$B = 4 \times 3.14 \times 10^{- 11} = 12.56 \times 10^{- 11} = 1.256 \times 10^{- 10} T .$

(iii) Magnetic moment assosciated with the electron

$M = IA = I ({πr}^{2}) = 10^{- 4} \times 3.14 \times (0.51 \times 10^{- 10})^{2} = 3.14 \times 0.2601 \times 10^{- 4} \times 10^{- 20}$

$M = 0.816 \times 10^{- 24} = 8.16 \times 10^{- 25} {Am}^{2}$

Some More Questions From Moving Charges And Magnetism Chapter

Two moving coil meters, M₁ and M₂ have the following particulars:
R₁ = 10 Ω N₁ = 30,
A₁ = 3.6 x 10^–3 m², B₁ = 0.25 T
R₂ = 14 Ω; N₂ = 42,
A₂ = 1.8 x 10^–3 m², B₂ = 0.50 T
(The spring constants k are identical for the two meters).
Determine the ratio of (a) current sensitivity and (b) voltage sensitivity of M₂ and M_1.

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