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The electron in a hydrogen atom revolves in an orbit of radius 0.5 Å, constituting a conventional current of 1.1 mA. Calculate the magnetic induction at an axial point 100 Å from the nucleus of theatom. |
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Answer» Solution :Data: `r=0.5 Å= 5 xx 10^(-11), I=1.1 xx 10^(-3) A, X=100 Å= 10^(-8)m, (mu_(0))/(4pi) = 10^(-7) T.m//A` The magnitude of the magnetic induction at an axial point of a current loop, `B=(mu_(0))/(4pi). (2IA)/(r^(2) + x^(2))^(3//2) ~= (mu_(0))/(4pi)(2IA)/(x^(3))(therefore x^(2) GT gt R^(2))` `=(mu_(0))/(4pi)(2I(pir^(2)))/(x^(3))` `=(10^(-7))(2(1.1 xx 10^(-3))(3.142)(5 xx 10^(-11))^(2))/(10^(-8))^(3)` `=3.142 xx 55 xx 10^(-8) = 1.728 xx 10^(-6)` T |
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