Verified Solution • 2025–26 Session moderate

QPage 276: Q6: HC Verma: Concepts of Physics - Volume 2 Permanent Magnets Page 276: Q6 Solution: Two short magnets of equal dipole moments M are fastened…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often plug the axial dipole formula $B=\frac{\mu_0}{4\pi}\frac{2M}{r^3}$ for each magnet even though the observation point lies 45° from each dipole’s axis; the correct field requires the angular components $B_r=\frac{\mu_0}{4\pi}\frac{2M\cos\theta}{r^3}$ and $B_\theta=\frac{\mu_0}{4\pi}\frac{M\sin\theta}{r^3}$, then vector‑add the two contributions. Using the axial magnitude and adding scalars leads to the wrong answer.

Two short magnets of equal dipole moments M are fastened perpendicularly at their centre in the Figure. The magnitude of the magnetic field at a distance d from the centre on the bisector of the right angle is

Multiple Choice Options

  • A μ04πMd3\frac{\mu_0}{4\pi} \frac{M}{d^3}
  • B μ04π2Md3\frac{\mu_0}{4\pi} \frac{\sqrt{2}M}{d^3}
  • C μ04π22Md3\frac{\mu_0}{4\pi} \frac{2\sqrt{2}M}{d^3}
  • D μ04π2Md3\frac{\mu_0}{4\pi} \frac{2M}{d^3}

Step-by-Step Verified Solution

**Answer:** (b) (__/4�)(�2M/d_). **Step-by-Step Explanation:** Two perpendicular dipoles give a resultant moment M�2. The field at a point on the bisector is (__/4�)(�2M/d_).

All Questions in this Chapter

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