A small bar magnet having a magnitic moment of `9xx10^(-3)Am^(-2)` is suspended at its centre of gravity by a light torsionless string at a distance of `10^(-2)m` vertically above a long, straight horizontal wire carrying a current of 1.0A from east to west. Find the frequency of oscillation of the magnet about its equilibrium position. The moment of inertia of the magnet is `6xx10^(-9)kgm^(2)`.(`H=3xx10^(-5)T`).
Text Solution
AI Generated Solution
To find the frequency of oscillation of the bar magnet about its equilibrium position, we will follow these steps:
### Step 1: Calculate the Magnetic Field due to the Current-Carrying Wire
The magnetic field \( B \) at a distance \( r \) from a long straight wire carrying current \( I \) is given by the formula:
\[
B = \frac{\mu_0 I}{2 \pi r}
\]
...
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