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A magnetic compass needle has a magnetic moment of 60 A·m² and is aligned along the geographical north at a certain location. If the torque acting on it is $1.2\times10^{-3}$ N·m and the horizontal component of Earth’s magnetic field at that place is 40 microtesla, what is the dip angle there?

1. 30°
2. 60°
3. 45°
4. 15°

Correct answer: 60°

Solution

The torque on a magnetic dipole is $\tau=mB\sin\theta$. Since the needle is aligned along north, the field is horizontal in the usual setup, so $\tau=mB_H$. This gives the horizontal field, and using the given horizontal component with the total field relation $B_H=B\cos\delta$, the dip angle comes out to $60^\circ$.

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