Verified Solution • 2025–26 Session moderate

QPages 277-278: Q18: HC Verma: Concepts of Physics - Volume 2 Permanent Magnets Pages 277-278: Q18 Solution: A short magnet produces a deflection of 37� in a deflection…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently confuse the Tan-A position formula with the Tan-B position, incorrectly using a power of 3 for the distance term ($d^3$) instead of the correct power of 4 ($d^4$) in the denominator. Always verify that the field of a short dipole on its axial line is proportional to $1/d^3$, which leads to the $d^4$ term in the magnetometer equation after accounting for the geometry.

A short magnet produces a deflection of 37� in a deflection magnetometer in Tan-A position when placed at a separation of 10 cm from the needle. Find the ratio of the magnetic moment of the magnet to the earth's horizontal magnetic field.

Step-by-Step Verified Solution

**Answer:**

M/B_H = 3.75 \times 10^3 \text{ A m}^2/\text{T}

. **Step-by-Step Explanation:** Given:

\theta = 37�

,

d = 0.1 \text{ m}

.

\tan \theta = \frac{\mu_0}{4\pi} \frac{2Md}{(d^2 - l^2)^2 B_H}

. For short magnet,

l \ll d

.

0.75 = 10^{-7} \times \frac{2M \times 0.1}{(0.01)^2 B_H} = 10^{-7} \times \frac{0.2M}{10^{-4} B_H}

.

M/B_H = \frac{0.75 \times 10^{-4}}{2 \times 10^{-8}} = 3.75 \times 10^3

[citation:12].

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