Use (i) The Ampere’s law for H and (ii) Continuity of lines of B, to conclude that inside a bar magnet, (a) Lines of H run from the A-pole to S-pole, while (b) Lines of B must run from the S-pole to A-pole

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- in the given figure let C be the amperian loop then,

 

${\int }_Q^{p}H.dl$ = ${\int }_Q^p\frac{B}{m_0}$ .dl

 between B and dl is less than 90 degree so

${\int }_Q^{p}H.dl$ = . ${\int }_Q^p\frac{B}{{\mu }_0}$ dl>0

So magnetic field from south pole to north pole inside the bar magnet

So according to ampere law

${\int }_{pQp}^{}H.dl$ =0

 

${\int }_{pQp}^{}H.dl$ = $\int H.dl$  +  $\int H.dl$  = 0

${\int }_Q^{p}H.dl$ >0, so ${\int }_p^{Q}H.dl$ <0,

It will be so if the angle between H and dl is more than 90 degree so cos $\theta$  is negative . it means H must run from north to south pole .

<p>This is a Short Answer Type Questions as classified in NCERT Exemplar</p><p><strong>Explanation-</strong> in the given figure let C be the amperian loop then, </p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1743158268phpYtpdGQ_480x360.jpeg" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1743158268phpYtpdGQ.jpeg" alt="" width="232" height="105"></picture></div><div>&nbsp;</div></div><p><span contenteditable="false"> <math> <msubsup> <mo>∫</mo> <mrow> <mi>Q</mi> </mrow> <mrow> <mi>p</mi> </mrow> </msubsup> <mrow> <mrow> <mi>H</mi> <mo>.</mo> <mi>d</mi> <mi>l</mi> </mrow> </mrow> </math> </span>= <span contenteditable="false"> <math> <msubsup> <mo>∫</mo> <mrow> <mi>Q</mi> </mrow> <mrow> <mi>p</mi> </mrow> </msubsup> <mrow> <mrow> <mfrac> <mrow> <mrow> <mi>B</mi> </mrow> </mrow> <mrow> <mrow> <msub> <mrow> <mrow> <mi>m</mi> </mrow> </mrow> <mrow> <mrow> <mn>0</mn> </mrow> </mrow> </msub> </mrow> </mrow> </mfrac> </mrow> </mrow> </math> </span>.dl</p><p>&nbsp;between B and dl is less than 90 degree so</p><p><span contenteditable="false"> <math> <msubsup> <mo>∫</mo> <mrow> <mi>Q</mi> </mrow> <mrow> <mi>p</mi> </mrow> </msubsup> <mrow> <mrow> <mi>H</mi> <mo>.</mo> <mi>d</mi> <mi>l</mi> </mrow> </mrow> </math> </span>= .<span contenteditable="false"> <math> <msubsup> <mo>∫</mo> <mrow> <mi>Q</mi> </mrow> <mrow> <mi>p</mi> </mrow> </msubsup> <mrow> <mrow> <mfrac> <mrow> <mrow> <mi>B</mi> </mrow> </mrow> <mrow> <mrow> <msub> <mrow> <mrow> <mi>μ</mi> </mrow> </mrow> <mrow> <mrow> <mn>0</mn> </mrow> </mrow> </msub> </mrow> </mrow> </mfrac> </mrow> </mrow> </math> </span>dl&gt;0</p><p>So magnetic field from south pole to north pole inside the bar magnet</p><p>So according to ampere law</p><p><span contenteditable="false"> <math> <msubsup> <mo>∫</mo> <mrow> <mi>p</mi> <mi>Q</mi> <mi>p</mi> </mrow> <mrow></mrow> </msubsup> <mrow> <mrow> <mi>H</mi> <mo>.</mo> <mi>d</mi> <mi>l</mi> </mrow> </mrow> </math> </span>=0</p><p>&nbsp;</p><p><span contenteditable="false"> <math> <msubsup> <mo>∫</mo> <mrow> <mi>p</mi> <mi>Q</mi> <mi>p</mi> </mrow> <mrow></mrow> </msubsup> <mrow> <mrow> <mi>H</mi> <mo>.</mo> <mi>d</mi> <mi>l</mi> </mrow> </mrow> </math> </span>=<span contenteditable="false"> <math> <mo>∫</mo> <mrow> <mrow> <mi>H</mi> <mo>.</mo> <mi>d</mi> <mi>l</mi> </mrow> </mrow> </math> </span>&nbsp;+&nbsp; <span contenteditable="false"> <math> <mo>∫</mo> <mrow> <mrow> <mi>H</mi> <mo>.</mo> <mi>d</mi> <mi>l</mi> </mrow> </mrow> </math> </span>&nbsp;= 0</p><p><span contenteditable="false"> <math> <msubsup> <mo>∫</mo> <mrow> <mi>Q</mi> </mrow> <mrow> <mi>p</mi> </mrow> </msubsup> <mrow> <mrow> <mi>H</mi> <mo>.</mo> <mi>d</mi> <mi>l</mi> </mrow> </mrow> </math> </span>&gt;0, so <span contenteditable="false"> <math> <msubsup> <mo>∫</mo> <mrow> <mi>p</mi> </mrow> <mrow> <mi>Q</mi> </mrow> </msubsup> <mrow> <mrow> <mi>H</mi> <mo>.</mo> <mi>d</mi> <mi>l</mi> </mrow> </mrow> </math> </span>&lt;0, </p><p>It will be so if the angle between H and dl is more than 90 degree so cos<span contenteditable="false"> <math> <mi>θ</mi> </math> </span>&nbsp;is negative . it means H must run from north to south pole .</p>

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