Final answer:
The direction of the magnetic field produced by a short bar magnet can be determined using the right-hand rule, with the field lines pointing away from the south pole and towards the north pole. The magnitude of the magnetic field is 2×10⁻⁶ times the current flowing through the magnet.
Step-by-step explanation:
To determine the direction and magnitude of the magnetic field produced by a short bar magnet at a distance of 10 cm from its center on the axis, we can use the formula for the magnetic field produced by a current loop:
B = (μ₀I)/(2πr)
where B is the magnetic field, μ₀ is the permeability of free space (4π×10⁻⁷ Tm/A), I is the current, and r is the distance from the center of the loop.
Since a short bar magnet can be approximated as a magnetic dipole, we can calculate the current using the formula for magnetic moment:
m = IA
where m is the magnetic moment, I is the current, and A is the area enclosed by the loop. In this case, the area is the cross-sectional area of the magnet.
Given that the magnetic moment is 0.48 JT⁻¹ and the distance is 10 cm (0.1 m), we can rearrange the formula for magnetic moment to solve for I:
I = m/A
Substituting the given values, we have:
I = 0.48 JT⁻¹ / A
Now we can substitute the calculated value of I into the formula for the magnetic field:
B = (4π×10⁻⁷ Tm/A × I) / (2π(0.1 m))
B = (2×10⁻⁷Tm/A × I) / (0.1 m)
B = 2×10⁻⁶ I T
Therefore, the direction of the magnetic field is determined by the right-hand rule, with the field lines pointing away from the south pole of the magnet and towards the north pole. The magnitude of the magnetic field is 2×10⁻⁶ times the current flowing through the magnet.