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A short bar magnet has a magnetic moment of 0.48JT⁻¹. Give the direction and magnitude of the magnetic field produced by the magnet at a distance of 10 cm from the centre of the magnet on the axis

User Nandish
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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.

User Ghostwriter
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