Ncert Exercises & Solutions

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)?

Magnetic moment,

M = 1.5 {JT}^{- 1}

Magnetic field strength, B=0.22 T

θ_{1} = 0 °

(i)Initial angle between the axis and the magnetic field,

θ_{1} = 0 °

Final angle between the axis and the magnetic field,

θ_{2} = 90 °

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

W = - M B (\cos θ_{2} - \cos θ_{1})

= 1.5 \times 0.22 (\cos 90 ° - \cos 0 °)

= 0.33 (0 - 1)

= 0.33 J

(ii) Initial angle between the axis and the magnetic field,

θ_{1} = 0 °

Final angle between the axis and the magnetic field,

θ_{2} = 180 °

The work required to make the magnetic moment opposite to the direction of magnetic field is given as:

W = - M B (\cos θ_{2} - \cos θ_{1})

= - 1.5 \times 0.22 (\cos 180 - \cos 0 °)

= - 0.33 (- 1 - 1)

= 0.66 J

(b) For case (i)

θ = θ_{2} = 90 °

∴ T o r q u e, τ = M B \sin θ

= 1.5 \times 0.22 \sin 90 °

= 0.33 J

For case (ii)

θ = θ_{2} = 180 °

∴ T o r q u e, τ = M B \sin θ

= M B \sin 180 ° = 0 J

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