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The electron in a H(2) atom circles ...

The electron in ` a H_(2) ` atom circles around the proton with a speed of` 2.18 xx 10^(6) ms^(-1) ` in an orbit of radius `5.3 xx 10^(-11)m`. Calculate (i) the equivalent current (ii) magnetic field produced at the proton.

Text Solution

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Charge , ` e = 1.6 xx 10^(-19) C ` and
` mu_(0) = 4 pi xx 10^(-7) T mA^(-1)`
`v = 2.18 xx 10^(6) ms^(-1)`
` r = 5.3 xx 10^(-11) m `
` e = 1.6 xx 10^(-19) C `
Time period , `T = (2pi r)/v`
` = (2pi xx 5.3 xx 10^(-11))/(2.18 xx 10^(6))`
`= 1.528 xx 10^(-16)s`
Equivalent current , `I = "Charge"/"time" = e/T`
` = (1.6 xx 10^(-19))/(1.528 xx 10^(16))`
` I = 1.05 xx 10^(-3) A`
(ii) Field at proton due to orbiting electron is
`B = (mu_(0)I)/(2 r) = (4 pi xx 10^(-7) xx 1.05 xx 10^(-3))/(2 xx 5.3 xx 10^(11)) `
` = 12.4 T `
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