Question 13

The magnetic moment of a bar magnet is $$0.5$$ Am$$^2$$. It is suspended in a uniform magnetic field of $$8 \times 10^{-2}$$ T. The work done in rotating it from its most stable to most unstable position is:

$$U = -MB \cos\theta$$

$$\text{Most stable position (parallel): } \theta_1 = 0^\circ$$

$$\text{Most unstable position (anti-parallel): } \theta_2 = 180^\circ$$

$$W = \Delta U = -MB(\cos\theta_2 - \cos\theta_1)$$

$$\implies W = -MB(\cos 180^\circ - \cos 0^\circ) = -MB(-1 - 1) = 2MB$$

$$\implies W = 2 \times 0.5 \times (8 \times 10^{-2}) = 8 \times 10^{-2}\text{ J}$$

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