Field of a Short Magnetic Dipole
At distance r much larger than the magnet size, a dipole m produces B = (mu_0/4pi r^3)[3(m dot r_hat)r_hat - m]. The axial magnitude is twice the equatorial magnitude at the same r.
Why this shows up in the exam
Fields of short bar magnets · Vector addition of fields from multiple dipoles · Estimating distance dependence of magnetic fields
Learn the idea
Far from a small magnet, its field falls as the inverse cube of distance and depends on direction. A bar magnet is strongest along its axis and weaker in the broadside equatorial direction. Far away, its size no longer matters separately; only its dipole moment controls the field.
🧠 Memory hook: Axial is two, equatorial is one, and both fade as r cubed.
Get this one clearly and it pays off every single time it shows up in the paper. 🎯
Formulas & facts to keep ready
- B_axial = (mu₀/(4 pi))(2m/r³) — far-field magnitude on the dipole axis
- B_equatorial = (mu₀/(4 pi))(m/r³) — far-field magnitude on the equatorial line, opposite to m
- B_net = vector sum of all B_i — superposition for several magnets
- V_m,axial = (mu₀/(4 pi)) m/r² — magnetic scalar potential on the dipole axis in the pole-model convention
How to approach it
- 1Mark each point as axial, equatorial, or general
- 2Compute each field with its direction
- 3Add vectors before taking the magnitude
Common slip-ups that cost marks
- •Using an inverse-square law
- •Ignoring the opposite direction on the equatorial line
- •Applying short-dipole formulas when distance is not large compared with magnet size
🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.
Original chapter practice
Original questions for this chapter, not past-paper questions or an exact mapping to this individual concept.
A charge of 2 microC moves perpendicular to a 3 T magnetic field at 4 x 10^5 m/s. Find the magnetic force.
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