A bar magnet of magnetic moment M and moment of inertia 1 (about centre, perpendicular to length) is cut into two equal pieces, perpendicular to length. Let T be the period of oscillations of the original magnet about an axis through the mid-point, perpendicular to length, in a magnetic field B. What would be the similar period T for each piece?

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alok kumar singh | Contributor-Level 10

4 months ago

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

Explanation- I= moment of inertia of the bar magnet

m= mass of bar magnet

l= length of magnet about an any passing through its centre and perpendicular to its

length

M= magnetic moment of the magnet

B= uniform magnetic field in which magnet Is oscillating

T= 2 $π \sqrt[]{\frac{I}{M B}}$

I= ml²/12

When magnet is cut into two equal pieces

I¹= $\frac{m {(l / 2)}^{2}}{2 (12)}$ = ml²/96= I/8

Magnetic dipole moment M^’= M/2

Its time period oscillation is T’= 2 $π \sqrt[]{\frac{I}{M B}}$ = 2 $π \sqrt[]{\frac{I / 8}{(M / 2) B}}$ = $\frac{2 π}{2} \sqrt[]{\frac{I}{M B}}$

This is a Short Answer Type Question as classified in NCERT ExemplarExplanation- I= moment of inertia of the bar magnetm= mass of bar magnetl= length of magnet about an any passing through its centre and perpendicular to itslengthM= magnetic moment of the magnetB= uniform magnetic field in which magnet Is oscillatingT= 2 <math> <mi>π</mi> <mroot> <mrow> <mrow> <mfrac> <mrow> <mrow> <mi>I</mi> </mrow> </mrow> <mrow> <mrow> <mi>M</mi> <mi>B</mi> </mrow> </mrow> </mfrac> </mrow> </mrow> <mrow></mrow> </mroot> </math> I= ml2/12When magnet is cut into two equal piecesI1= <math> <mfrac> <mrow> <mrow> <mi>m</mi> <msup> <mrow> <mrow> <mo> (</mo> <mi>l</mi> <mo>/</mo> <mn>2</mn> <mo>)</mo> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> <mi> </mi> <mo> (</mo> <mn>12</mn> <mo>)</mo> </mrow> </mrow> </mfrac> </math> = ml2/96= I/8Magnetic dipole moment M’= M/2Its time period oscillation is T’= 2 <math> <mi>π</mi> <mroot> <mrow> <mrow> <mfrac> <mrow> <mrow> <mi>I</mi> </mrow> </mrow> <mrow> <mrow> <mi>M</mi> <mi>B</mi> </mrow> </mrow> </mfrac> </mrow> </mrow> <mrow></mrow> </mroot> </math> = 2 <math> <mi>π</mi> <mroot> <mrow> <mrow> <mfrac> <mrow> <mrow> <mi>I</mi> <mo>/</mo> <mn>8</mn> </mrow> </mrow> <mrow> <mrow> <mo> (</mo> <mi>M</mi> <mo>/</mo> <mn>2</mn> <mo>)</mo> <mi>B</mi> </mrow> </mrow> </mfrac> </mrow> </mrow> <mrow></mrow> </mroot> </math> = <math> <mfrac> <mrow> <mrow> <mn>2</mn> <mi>π</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mfrac> <mroot> <mrow> <mrow> <mfrac> <mrow> <mrow> <mi>I</mi> </mrow> </mrow> <mrow> <mrow> <mi>M</mi> <mi>B</mi> </mrow> </mrow> </mfrac> </mrow> </mrow> <mrow></mrow> </mroot> </math>

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