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The magnetic field at the center of a current carrying loop of radius 0.1 m is 5sqrt( 5) times that at a point along its axis. The distance of this point from the centre of the loop is |
| Answer» <html><body><p>0.<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a> m <br/>0.1 m <br/>0.05 m <br/>0.25 m </p>Solution :We know that,<br/>`(B_("centre"))/(B_("axis"))=(1+(x^(2))/(r2))^(3//2)`<br/>Given that, `B_("centre")=5sqrt(<a href="https://interviewquestions.tuteehub.com/tag/5-319454" style="font-weight:bold;" target="_blank" title="Click to know more about 5">5</a>)B_("axis")`<br/>`(B_("centre"))/(B_("axis"))=5sqrt(5)`<br/>`therefore 5sqrt(5)=[1+(x^(2))/((0.1)^(2))]^(3//2)`<br/>On squaring both sides, we get<br/>`25xx5 = [1+(x^(2))/((0.1)^(2))]`<br/>`root(3)(<a href="https://interviewquestions.tuteehub.com/tag/125-271180" style="font-weight:bold;" target="_blank" title="Click to know more about 125">125</a>)=1+(x^(2))/((0.1)^(2))`<br/>`<a href="https://interviewquestions.tuteehub.com/tag/rarr-1175461" style="font-weight:bold;" target="_blank" title="Click to know more about RARR">RARR</a> 0.01 + x^(2)=0.05`<br/>`rArr x^(2)=0.005 - 0.01`<br/>`rArr x^(2)=0.04`<br/>`rArr x = 0.2 m`</body></html> | |