Question

Figure below shows a small magnetized needle $A$ 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 magnetized needle $B$ .

Which configuration corresponds to the lowest potential energy among all the configurations shown?

Question Image

Which configuration corresponds to the lowest potential energy among all the configurations shown?

Accepted Answer

Figure below shows the direction of magnetic field of dipole $A$ and the magnetic moment of $B$ at different positions.

Potential energy of the configuration arises due to the potential energy of one dipole (say $B$ ) in the magnetic field of the other dipole $(A)$ . The magnitude of the magnetic field of dipole $A$ at any point on its normal bisector is given by

$B_{A} = \frac{μ_{0}}{4 π} \cdot \frac{m_{A}}{r^{3}}$ (on the normal bisector)

Also, the magnitude of the magnetic field of dipole $A$ at any point on its axis is given by

$B_{A} = \frac{μ_{0}}{4 π} \cdot \frac{2 m_{A}}{r^{3}}$ (on the axis)

Potential energy of the configuration at any point will be

$U = - {\vec{m}}_{B} \cdot {\vec{B}}_{A}$

(i) In configuration $A B_{1}, m_{B} ⊥ B_{A}$

$∴ U_{1} = - m_{B} B_{A} \cos 90^{\circ} = 0$

(ii) In configuration $A B_{2}, m_{B} ⊥ B_{A}$

$∴ U_{2} = - m_{B} B_{A} \cos 90^{\circ} = 0$

(iii) In configuration $A B_{3}, m_{B}$ is parallel to $B_{A}$

$∴ U_{3} = - m_{B} B_{A} \cos 0^{\circ} = - m_{B} \cdot \frac{μ_{0}}{4 π} \frac{m_{A}}{r^{3}}$

$= - \frac{μ_{0}}{4 π} \frac{m_{A} m_{B}}{r^{3}}$

(iv) In configuration $A B_{4}, m_{B}$ is antiparallel to $B_{A}$

$∴ U_{4} = - m_{B} B_{A} \cos 180^{\circ} = m_{B} \cdot \frac{μ_{0}}{4 π} \frac{2 m_{A}}{r^{3}}$

$= \frac{μ_{0}}{4 π} \cdot \frac{2 m_{A} m_{B}}{r^{3}}$

(v) In configuration $A B_{5}, m_{B}$ is antiparallel to $B_{A}$

$∴ U_{5} = - m_{B} B_{A} \cos 180^{\circ} = m_{B} \cdot \frac{μ_{0}}{4 π} \frac{m_{A}}{r^{3}}$

$= \frac{μ_{0}}{4 π} \cdot \frac{m_{A} m_{B}}{r^{2}}$

(vi) In configuration $A B_{6}, m_{B}$ is parallel to $B_{A}$

$U_{6} = - m_{B} B_{A} \cos 0^{\circ} = - m_{B} \frac{μ_{0}}{4 π} \cdot \frac{2 m_{A}}{r^{3}}$

$= - \frac{μ_{0}}{4 π} \cdot \frac{2 m_{A} m_{B}}{r^{3}}$

Obviously, the system has lowest potential energy in configuration $A B_{6}$ .

S L Arora Solutions for Chapter: Magnetism, Exercise 4: HOTS Problem on Higher Order Thinking Skills

S L Arora Physics Solutions for Exercise - S L Arora Solutions for Chapter: Magnetism, Exercise 4: HOTS Problem on Higher Order Thinking Skills

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