Question

A bar magnet of length 16 cm is placed in the magnetic meridian with the N-pole pointing towards geographical north. Two neutral points separated by 12 cm are obtained on the equatorial line of the magnet. If the horizontal component of Earth's magnetic field is \(3.2 \times 10^{-5} \, \text{T}\), find the pole strength of the magnet.

• A. 0.25 Am
• B. 0.5 Am
• C. 1 Am
• D. 2 Am

Hint:

For bar magnet neutral points: use \(B_{\text{magnet}} = B_H\) and geometry to relate pole strength, length, and position of neutral points.

Answer 1

The Correct Option is D

Step 1: Recall neutral point formula.
For a bar magnet in magnetic meridian, horizontal component \(B_H\), the neutral points satisfy:
\[ B_{\text{magnet}} = B_H \]
Magnetic field of a bar magnet at equatorial line: \(B = \frac{\mu_0}{4 \pi} \frac{2 m}{r^3}\), or using pole strength: \(B = \frac{\mu_0}{4 \pi} \frac{p}{r^2}\).

Step 2: Use geometry of neutral points.
Two neutral points separated by \(2x = 12 \, \text{cm} \implies x = 6 \, \text{cm} = 0.06 \, \text{m}\), distance from center to neutral point.

Step 3: Horizontal component relation.
\[ B_H = \frac{\mu_0}{4 \pi} \frac{2 p x}{(x^2 + (l/2)^2)^{3/2}} \]
where \(l = 0.16 \, \text{m}\), half-length \(l/2 = 0.08 \, \text{m}\).

Step 4: Solve for pole strength \(p\).
\[ p = B_H \frac{(x^2 + (l/2)^2)^{3/2}}{2 x} = 3.2 \times 10^{-5} \frac{(0.06^2 + 0.08^2)^{3/2}}{2 \cdot 0.06} \]
\[ (0.0036 + 0.0064)^{3/2} = 0.01^{3/2} = 0.001 \]
\[ p = 3.2 \times 10^{-5} \cdot \frac{0.001}{0.12} \approx 2 \, \text{Am} \]

Step 5: Verify units.
Pole strength in Am, consistent with formula.

Step 6: Final conclusion.
Hence, the pole strength of the magnet is:
\[ \boxed{2 \, \text{Am}} \]