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| 1. |
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. |
| Answer» <html><body><p></p>Solution :Data: `r=<a href="https://interviewquestions.tuteehub.com/tag/0-251616" style="font-weight:bold;" target="_blank" title="Click to know more about 0">0</a>.5 Å= 5 xx 10^(-11), I=1.1 xx 10^(-3) A, <a href="https://interviewquestions.tuteehub.com/tag/x-746616" style="font-weight:bold;" target="_blank" title="Click to know more about X">X</a>=100 Å= 10^(-<a href="https://interviewquestions.tuteehub.com/tag/8-336412" style="font-weight:bold;" target="_blank" title="Click to know more about 8">8</a>)m, (mu_(0))/(4pi) = 10^(-7) T.m//A` <br/> The magnitude of the magnetic induction at an axial point of a current loop, <br/> `B=(mu_(0))/(4pi). (2IA)/(r^(2) + x^(2))^(3//2) ~= (mu_(0))/(4pi)(2IA)/(x^(3))(therefore x^(2) <a href="https://interviewquestions.tuteehub.com/tag/gt-1013864" style="font-weight:bold;" target="_blank" title="Click to know more about GT">GT</a> gt R^(2))` <br/> `=(mu_(0))/(4pi)(2I(pir^(2)))/(x^(3))` <br/> `=(10^(-7))(2(1.1 xx 10^(-3))(3.142)(5 xx 10^(-11))^(2))/(10^(-8))^(3)` <br/> `=3.142 xx 55 xx 10^(-8) = 1.728 xx 10^(-6)` T</body></html> | |