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Worked Examples · Example 5.2

Q.Figure 5.4 shows a small magnetised needle P placed at a point O. The arrow shows the direction of its magnetic moment. The other arrows show different positions (and orientations of the magnetic moment) of another identical magnetised needle Q.

Figure 5.4 — Illustration for Example 5.2 — magnetised needles around a central dipole P.
Figure 5.4
(a) In which configuration the system is not in equilibrium?
(b) In which configuration is the system in
(i) stable, and
(ii) unstable equilibrium?
(c) Which configuration corresponds to the lowest potential energy among all the configurations shown?
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P's moment points up, so its field is upward on the axis (at Q4, Q6) and downward on the equator (at Q1, Q2, Q3, Q5). A needle is in equilibrium only when its moment is parallel (stable) or antiparallel (unstable) to the local field. (a) Q1 and Q2 are not in equilibrium. (b) Stable: Q3 and Q6; unstable: Q4 and Q5. (c) Q6 has the lowest potential energy.

Concept Understanding

Each needle is a magnetic dipole. Needle Q sits in the field B⃗P\vec{B}_P of the central needle P, whose moment points up. The orientation energy is U=−m⃗Q⋅B⃗P=−mBcos⁡θU=-\vec{m}_Q\cdot\vec{B}_P=-mB\cos\theta, where θ\theta is the angle between m⃗Q\vec{m}_Q and the local field.

  • On P's axis (top point Q4, bottom point Q6) the dipole field is parallel to m⃗P\vec{m}_P (points up), with magnitude Baxial=μ04π2mr3B_{\text{axial}}=\dfrac{\mu_0}{4\pi}\dfrac{2m}{r^3}.
  • On P's equator (Q1, Q2 near O, and Q3, Q5 on the circle) the field is antiparallel to m⃗P\vec{m}_P (points down), with magnitude Beq=μ04πmr3B_{\text{eq}}=\dfrac{\mu_0}{4\pi}\dfrac{m}{r^3}.

Zero torque (τ⃗=m⃗Q×B⃗P=0\vec{\tau}=\vec{m}_Q\times\vec{B}_P=0) requires m⃗Q\vec{m}_Q parallel or antiparallel to the local field. Parallel (θ=0\theta=0) is a potential-energy minimum, so stable; antiparallel (θ=180∘\theta=180^\circ) is a maximum, so unstable.

Step-by-Step Solution

NeedlePositionLocal B⃗P\vec{B}_Pm⃗Q\vec{m}_Qθ\thetaVerdict
Q1near O (equator)downright90∘90^\circtorque ≠0\neq 0 -> not in equilibrium
Q2near O (equator)downright90∘90^\circtorque ≠0\neq 0 -> not in equilibrium
Q3right (equator)downdown0∘0^\circstable

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