First slide

Magnetic dipole in a uniform magnetic field

Question

In an uniform field the magnetic needle completes 10 oscillations in 92 seconds. When a small magnet is placed in the magnetic meridian 10 cm due north of needle with north pole towards south completes 15 oscillations in 69 seconds. The magnetic moment of magnet (B_H = 0.3 G) is

Moderate

A

4.5 Am²
B

0.45 Am²
C

0.75 Am²
D

0.225 Am²

Solution

$\begin{array}{rcl} \frac{f^{1}}{f} & = & \sqrt{\frac{B^{1}}{B}}; \\ \frac{92}{10} x \frac{15}{69} & = & \sqrt{\frac{B^{1}}{0.3}}; \\ B^{1} & = & 1.2 G \\ B_{1} - B_{H} & = & 1.2; \\ B_{1} & = & 1.5 G \\ = & 1.5 \times 10^{- 4} T \\ \frac{μ_{0}}{4 π} \frac{2 M}{d^{3}} & = & B_{1}; \\ M & = & 0.75 {Am}^{2} \end{array}$

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