A short bar magnet has a magnetic moment of 0.4 `JT^(-1)` The magnitude of the magnetic field (in gauss) produced by the magnet at a distance of `20 cm` from the center of the magnet on the equatorial line of the magnet is

Answer

To find the magnitude of the magnetic field produced by a short bar magnet at a distance from its center on the equatorial line, we can use the formula for the magnetic field \( B \) at a distance \( r \) from the center of the magnet:

\[ B = \frac{\mu_0}{4\pi} \cdot \frac{2M}{r^3} \]

Where:
- \( B \) is the magnetic field in Tesla,
- \( \mu_0 \) is the permeability of free space, approximately \( 4\pi \times 10^{-7} \, \text{T m/A} \),
- \( M \) is the magnetic moment of the magnet in \( \text{JT}^{-1} \),
- \( r \) is the distance from the center of the magnet in meters.

Given:
- \( M = 0.4 \, \text{JT}^{-1} \)
- \( r = 20 \, \text{cm} = 0.2 \, \text{m} \)

Now, substituting the values into the formula:

1. Calculate \( r^3 \):
\[ r^3 = (0.2)^3 = 0.008 \, \text{m}^3 \]

2. Substitute \( M \) and \( r^3 \) into the formula:
\[ B = \frac{4\pi \times 10^{-7}}{4\pi} \cdot \frac{2 \times 0.4}{0.008} \]
Simplifying gives:
\[ B = 10^{-7} \cdot \frac{0.8}{0.008} \]

3. Calculate \( \frac{0.8}{0.008} \):
\[ \frac{0.8}{0.008} = 100 \]

4. Thus, the magnetic field \( B \) in Tesla is:
\[ B = 10^{-7} \cdot 100 = 10^{-5} \, \text{T} \]

5. Convert \( B \) from Tesla to Gauss:
\[ 1 \, \text{T} = 10^4 \, \text{Gauss} \]
Therefore,
\[ B = 10^{-5} \, \text{T} \times 10^4 \, \text{Gauss/T} = 0.1 \, \text{Gauss} \]

Thus, the magnitude of the magnetic field produced by the magnet at a distance of 20 cm from its center on the equatorial line is **0.1 Gauss**.