An iron bar of length $L$ has magnetic moment $M$ . It is bent at the middle of its length such that the two arms make an angle $60^{?}$  with each other. The magnetic moment of this new magnet is :

Correct Option - 2


Detailed Solution:

$M=ml$

$\Delta I=2\frac{l}{2}sin30^{?}$

$=\frac{l}{2}$

$M^{\prime }=mI/2$

$=M/2$

<div><div><picture><source srcset="" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/1754999158phpqyYfpL.jpeg" alt="" width="241" height="55" loading="lazy"></picture></div></div><p><span contenteditable="false"> <math> <mrow> <mi>M</mi> <mo>=</mo> <mi>m</mi> <mi>l</mi> </mrow> </math> </span></p><div><div><picture><source srcset="" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/1754999196php1pLeIg.jpeg" alt="" width="220" height="193" loading="lazy"></picture></div></div><p><span contenteditable="false"> <math> <mrow><mi>Δ</mi><mi>I</mi><mo>=</mo><mn>2</mn><mfrac><mrow><mi>l</mi></mrow><mrow><mn>2</mn></mrow></mfrac>sin<mn>3</mn><msup><mrow><mn>0</mn></mrow><mrow><mo>? </mo></mrow></msup></mrow> </math> </span></p><p><span contenteditable="false"> <math> <mrow> <mo>=</mo> <mfrac> <mrow> <mi>l</mi> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> </mrow> </math> </span></p><p><span contenteditable="false"> <math> <mrow> <msup> <mi>M</mi> <mo>′</mo> </msup> <mo>=</mo> <mi>m</mi> <mi>I</mi> <mo>/</mo> <mn>2</mn> </mrow> </math> </span></p><p><span contenteditable="false"> <math> <mrow> <mo>=</mo> <mi>M</mi> <mo>/</mo> <mn>2</mn> </mrow> </math> </span></p>

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