A compass originally points North; at this location the horizontal component of

Question

A compass originally points North; at this location the horizontal component of the Earth's magnetic field has a magnitude of 2e-5 T. A bar magnet is aligned East-West, pointing at the center of the compass. When the center of the magnet is 0.29 m from the center of the compass, the compass deflects 70 degrees. What is the magnetic dipole moment of the bar magnet?

The answer of course will be in A*m^2. I searched chegg and found no good and correct answers, so please don't copy paste someone elses answer:) Thanks

Explanation / Answer

referring to his formula, the magnetic dipole m is parallel to r since the magnet is pointed directly at the compass. Thus we get for the field due to the magnet:

B = (?0/4?) 2m / r2 = 1e-7*(2/0.292)m and points in the same direction as m namely East-West.

The Earth's field (BE) and the magnet's field are at 90 deg to each other (one is N-S and the other is E-W). Hence the deflection ? of the compass (which would align itself with the nett field) is given by tan? = B/BE (draw a picture to convince yourself) giving B = BE tan? =55e-6 Tesla as the field due to the bar magnet.

Substituting back in the expression for field to a dipole, m=23.12 A-m2 is the dipole moment of the magnet (answer). As always, this is only intended to be a guide, and you should check the arithmetic.

pls rate me first. thanks alot   

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