A short bar magnet with magnetic dipole moment 1.6\text{ A}\cdot\text{m}^2 is placed along… — JAMB UTME

A short bar magnet with magnetic dipole moment $1.6\text{ A}\cdot\text{m}^2$ is placed along the magnetic meridian with its north pole pointing towards the Earth's magnetic south pole. A neutral point is located on the axial line of the magnet at a distance of $0.2\text{ m}$ from its center. What is the magnitude of the horizontal component of the Earth's magnetic field at this location, in microtesla ( $\mu\text{T}$ )? (Take $\frac{\mu_0}{4\pi} = 10^{-7}\text{ T}\cdot\text{m}\cdot\text{A}^{-1}$ )

Answer: 40 μT

Answer

The magnitude of the horizontal component of Earth's magnetic field at this location is 40 μT.

At a neutral point, the horizontal component of Earth's magnetic field is equal in magnitude and opposite in direction to the magnetic field generated by the bar magnet. For a magnet aligned with its north pole pointing south, neutral points lie on its axial line at distance

d

. Using

B_h = \frac{\mu_0}{4\pi} \frac{2M}{d^3}

with

M = 1.6\text{ A}\cdot\text{m}^2

and

d = 0.2\text{ m}

yields

B_h = 4.0 \times 10^{-5}\text{ T} = 40\ \mu\text{T}

Step-by-Step Solution

Determine the condition for the neutral point

B_{\text{axial}} = B_h

When a magnet's north pole points south, its axial magnetic field opposes Earth's horizontal field, creating neutral points along the axis where the magnetic fields cancel out completely.

Apply the short bar magnet formula for field along the axial line

B_h = \frac{\mu_0}{4\pi} \frac{2M}{d^3}

The magnetic field produced at an axial point at distance

d

from the center of a short bar magnet of magnetic moment

M

is given by this formula.

Substitute the given numerical parameters

B_h = 10^{-7} \times \frac{2 \times 1.6}{(0.2)^3}

Substituting

M = 1.6\text{ A}\cdot\text{m}^2

d = 0.2\text{ m}

, and

\frac{\mu_0}{4\pi} = 10^{-7}\text{ T}\cdot\text{m}\cdot\text{A}^{-1}

into the field equation.

Calculate the magnitude of the horizontal field component in microtesla

B_h = 4.0 \times 10^{-5}\text{ T} = 40\ \mu\text{T}

Dividing

3.2 \times 10^{-7}

8 \times 10^{-3}

gives

4 \times 10^{-5}\text{ T}

, which converts to

40\ \mu\text{T}

Key Concept

Neutral points created by a bar magnet aligned with Earth's magnetic meridian

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