The magnetic moment of a bar magnet is 1 joule per tesla. When bar magnet makes $30^{0}$ with the field, we need a external torque of $3 N m$ to hold it. Find the work required to turn it by further $90^{0}$ .

6.3 J

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8.1 J

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4.9 J

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19.5 J

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Solution

The correct option is B $8.1 J$
$(τ | = (M \times B | = M B sin θ = M B sin 30^{0} = \frac{M B}{2} = 3$
$W_{e x t} = U_{f} - U_{i} = - M B cos θ_{f} + M B cos θ_{i} = 6 (cos 30^{0} - cos 120^{0}) = 3 (1 + \sqrt{3}) = 3 (2.7) = 8.1 J$

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The magnetic moment of a bar magnet is 1 joule per tesla. When bar magnet makes 300 with the field, we need a external torque of 3 Nm to hold it. Find the work required to turn it by further 900.

The correct option is B 8.1 J(τ| = (M×B| = MBsinθ =MBsin300=MB2 = 3 Wext = Uf − Ui = −MBcos θf +MB cosθi = 6( cos300 − cos1200) = 3(1 + √3) = 3(2.7) = 8.1 J

The magnetic moment of a bar magnet is 1 joule per tesla. When bar magnet makes $30^{0}$ with the field, we need a external torque of $3 N m$ to hold it. Find the work required to turn it by further $90^{0}$ .

The correct option is B $8.1 J$
$(τ | = (M \times B | = M B sin θ = M B sin 30^{0} = \frac{M B}{2} = 3$
$W_{e x t} = U_{f} - U_{i} = - M B cos θ_{f} + M B cos θ_{i} = 6 (cos 30^{0} - cos 120^{0}) = 3 (1 + \sqrt{3}) = 3 (2.7) = 8.1 J$