First slide

Magnetic dipole in a uniform magnetic field

Question

A short magnet oscillates in vibration magnetometer with a frequency 10 Hz where horizontal component of earth's magnetic field is 12 $μ$ T. A downward current of 15 A is established in the vertical wire placed 20 cm west of the magnet. New frequency is

Difficult

A

4Hz
B

2.5Hz
C

9Hz
D

5Hz

Solution

$\begin{array}{rcl} B_{1} & = & \frac{μ_{0} i}{2 πr} = 150 x 10^{- 7} = 15 μT \\ B & = & B_{1} - B_{H} = (15 - 3) μT = 12 μT \\ \frac{f^{1}}{f} & = & \sqrt{\frac{B_{H}}{B}} = \sqrt{\frac{3 μT}{12 μT}} = \frac{1}{2}; \\ f^{1} & = & \frac{10}{2} = 5 Hz \end{array}$

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