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)?

Question

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Answer 1

Here $m = 1.5 J / T, B = 0.22 T$
(a)
$W = - m B (c o s θ_{2} - c o s θ_{1})$

(i)
$θ_{1} = 0^{\circ}$ (along the field)
$θ_{2} = 9 0^{\circ}$ (perpendicular to the field)
$W = - 0.33 (0 - 1) J = 0.33 J$

(ii)
$θ_{1} = 0^{\circ}, θ_{2} = 18 0^{\circ}$
$W = - 1.5 \times 0.22 (c o s 18 0^{\circ} - c o s 0^{\circ}) = 0.66 J$

(b)
Torque $= m B s i n θ$
(i)
$θ = 9 0^{\circ}$
$⟹ τ = 0.33 N m$
(ii)
$τ = m B s i n 18 0^{\circ}$
$= 0 N m$

Answer 2

Here

m = 1.5 J / T, B = 0.22 T

(a)

W = - m B (c o s θ_{2} - c o s θ_{1})

(i)

θ_{1} = 0^{\circ}

(along the field)

θ_{2} = 9 0^{\circ}

(perpendicular to the field)

W = - 0.33 (0 - 1) J = 0.33 J

(ii)

θ_{1} = 0^{\circ}, θ_{2} = 18 0^{\circ}

W = - 1.5 \times 0.22 (c o s 18 0^{\circ} - c o s 0^{\circ}) = 0.66 J

(b)

Torque

= m B s i n θ

(i)

θ = 9 0^{\circ}

⟹ τ = 0.33 N m

(ii)

τ = m B s i n 18 0^{\circ}

= 0 N m