5.7 A bar magnet of magnetic moment 1.5 J T^–1 lies aligned with the direction of a uniform magnetic field of 0.22 T.

(a) What is the amount of work required by an external torque to turn the magnet so as to align its magnetic moment: (i) normal to the field direction, (ii) opposite to the field direction?

(b) What is the torque on the magnet in cases (i) and (ii)?

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Payal Gupta | Contributor-Level 10

5 months ago

5.7 Magnetic moment, M = 1.5 J/T

Magnetic field strength, B = 0.22 T

Initial angle between the axis and the magnetic field, $θ_{1}$ = 0 $°$ and the final angle, $θ_{2}$ = 90 $°$

Work required to make the magnetic moment normal to the direction of the magnetic field is given as:

W = -MB( cos $θ_{2} -$ cos $θ_{1}$ ) = - 1.5 $\times 0.22 (\cos ? 90 °$ - cos 0 $°$ ) = 0.33 J

Initial angle between the axis and the magnetic field, $θ_{1}$ = 0 $°$ and the final angle, $θ_{2}$ = 180

Work required to make the magnetic moment normal to the direction of the magnetic field is given as:

W = -MB( cos $θ_{2} -$ cos $θ_{1}$ ) = - 1.5 $\times 0.22 (\cos ? 180 °$ - cos 0 $°$ ) = 0.33 J

For case

Torque $τ = M B \sin ? θ$ , here $θ$ = $θ_{2}$ = 90 $°$

= 1.5 $\times 0.22 \sin ? 90 °$ = 0.33 J

Torque $τ = M B \sin ? θ$ , here $θ$ = $θ_{2}$ = 180 $°$

= 1.5 $\times 0.22 \sin ? 180 °$ =

...more

5.7 Magnetic moment, M = 1.5 J/T

Magnetic field strength, B = 0.22 T

Initial angle between the axis and the magnetic field, $θ_{1}$ = 0 $°$ and the final angle, $θ_{2}$ = 90 $°$

Work required to make the magnetic moment normal to the direction of the magnetic field is given as:

W = -MB( cos $θ_{2} -$ cos $θ_{1}$ ) = - 1.5 $\times 0.22 (\cos ? 90 °$ - cos 0 $°$ ) = 0.33 J

Initial angle between the axis and the magnetic field, $θ_{1}$ = 0 $°$ and the final angle, $θ_{2}$ = 180

Work required to make the magnetic moment normal to the direction of the magnetic field is given as:

W = -MB( cos $θ_{2} -$ cos $θ_{1}$ ) = - 1.5 $\times 0.22 (\cos ? 180 °$ - cos 0 $°$ ) = 0.33 J

For case

Torque $τ = M B \sin ? θ$ , here $θ$ = $θ_{2}$ = 90 $°$

= 1.5 $\times 0.22 \sin ? 90 °$ = 0.33 J

Torque $τ = M B \sin ? θ$ , here $θ$ = $θ_{2}$ = 180 $°$

= 1.5 $\times 0.22 \sin ? 180 °$ = 0 J

less

5.7 Magnetic moment, M = 1.5 J/TMagnetic field strength, B = 0.22 TInitial angle between the axis and the magnetic field, <math> <msub> <mrow> <mrow> <mi>θ</mi> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> </msub> </math> = 0 <math> <mo>°</mo> </math> and the final angle, <math> <msub> <mrow> <mrow> <mi>θ</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msub> </math> = 90 <math> <mo>°</mo> </math> Work required to make the magnetic moment normal to the direction of the magnetic field is given as:W = -MB( cos <math> <msub> <mrow> <mrow> <mi> </mi> <mi>θ</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msub> <mo>-</mo> <mi> </mi> </math> cos <math> <msub> <mrow> <mrow> <mi>θ</mi> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> </msub> </math> ) = - 1.5 <math> <mo>×</mo> <mn>0.22</mn> <mi> </mi> <mo>(</mo> <mrow> <mrow> <mi mathvariant="normal">cos</mi> </mrow> <mo>?</mo> <mrow> <mn>90</mn> <mo>°</mo> </mrow> </mrow> </math> - cos 0 <math> <mo>°</mo> </math> ) = 0.33 JInitial angle between the axis and the magnetic field, <math> <msub> <mrow> <mrow> <mi>θ</mi> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> </msub> </math> = 0 <math> <mo>°</mo> </math> and the final angle, <math> <msub> <mrow> <mrow> <mi>θ</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msub> </math> = 180Work required to make the magnetic moment normal to the direction of the magnetic field is given as:W = -MB( cos <math> <msub> <mrow> <mrow> <mi> </mi> <mi>θ</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msub> <mo>-</mo> <mi> </mi> </math> cos <math> <msub> <mrow> <mrow> <mi>θ</mi> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> </msub> </math> ) = - 1.5 <math> <mo>×</mo> <mn>0.22</mn> <mi> </mi> <mo>(</mo> <mrow> <mrow> <mi mathvariant="normal">cos</mi> </mrow> <mo>?</mo> <mrow> <mn>180</mn> <mo>°</mo> </mrow> </mrow> </math> - cos 0 <math> <mo>°</mo> </math> ) = 0.33 JFor caseTorque <math> <mi>τ</mi> <mo>=</mo> <mi>M</mi> <mi>B</mi> <mrow> <mrow> <mi mathvariant="normal">sin</mi> </mrow> <mo>?</mo> <mrow> <mi>θ</mi> </mrow> </mrow> </math> , here <math> <mi>θ</mi> </math> = <math> <msub> <mrow> <mrow> <mi>θ</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msub> </math> = 90 <math> <mo>°</mo> </math> = 1.5 <math> <mo>×</mo> <mn>0.22</mn> <mrow> <mrow> <mi mathvariant="normal">sin</mi> </mrow> <mo>?</mo> <mrow> <mn>90</mn> <mo>°</mo> </mrow> </mrow> </math> = 0.33 JTorque <math> <mi>τ</mi> <mo>=</mo> <mi>M</mi> <mi>B</mi> <mrow> <mrow> <mi mathvariant="normal">sin</mi> </mrow> <mo>?</mo> <mrow> <mi>θ</mi> </mrow> </mrow> </math> , here <math> <mi>θ</mi> </math> = <math> <msub> <mrow> <mrow> <mi>θ</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msub> </math> = 180 <math> <mo>°</mo> </math> = 1.5 <math> <mo>×</mo> <mn>0.22</mn> <mrow> <mrow> <mi mathvariant="normal">sin</mi> </mrow> <mo>?</mo> <mrow> <mn>180</mn> <mo>°</mo> </mrow> </mrow> </math> = 0 J

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