Q.(a) A current-carrying circular loop lies on a smooth horizontal plane. Can a uniform magnetic field be set up in such a manner that the loop turns around itself (i.e., turns about the vertical axis).

(a) Turning about the vertical axis

A uniform magnetic field exerts a torque $\vec{\tau} = \vec{m} \times \vec{B}$ on a current loop, where $\vec{m}$ is the magnetic moment. This torque always tries to align $\vec{m}$ with $\vec{B}$, not to spin the loop about a vertical axis. For the loop to turn around itself (vertical axis), the torque must have a vertical component. But in a uniform field, $\vec{\tau}$ is perpendicular to both $\vec{m}$ and $\vec{B}$ — it lies in the horizontal plane. Hence, no uniform field can make the loop rotate about the vertical axis. …

(a) Can a uniform magnetic field make the loop turn about a vertical axis?

Concept and intuition:

A current-carrying loop in a uniform magnetic field experiences no net force — the forces on opposite sides cancel. What it does experience is a torque, given by $\vec{\tau} = \vec{m} \times \vec{B}$, where $\vec{m} = I \vec{A}$ is the magnetic moment of the loop (direction given by the right-hand rule). This torque tries to align $\vec{m}$ with $\vec{B}$, not to spin the loop continuously. For the loop to “turn around itself” (rotate about a vertical axis), the torque would need to have a vertical component. But in a uniform field, the torque is always perpendicular to both $\vec{m}$ and $\vec{B}$, and if the field is horizontal, the torque is horizontal — it cannot produce a vertical rotation. If the field is vertical, the torque is zero because $\vec{m}$ and $\vec{B}$ are parallel or antiparallel. So no uniform field can cause continuous rotation about a vertical axis.

Step-by-step reasoning:

1. Magnetic moment direction: For a circular loop lying on a smooth horizontal plane, the area vector $\vec{A}$ (and hence $\vec{m} = I\vec{A}$) is vertical — either upward or downward depending on current direction. So $\vec{m}$ is along the vertical axis.

2. Torque in a uniform field: The torque is $\vec{\tau} = \vec{m} \times \vec{B}$. For $\vec{\tau}$ to have a vertical component (to turn the loop about the vertical axis), $\vec{m}$ and $\vec{B}$ must have horizontal components that are not parallel. But $\vec{m}$ is purely vertical. The cross product of a vertical vector with any vector $\vec{B}$ gives a horizontal torque — never vertical. So the loop cannot rotate about the vertical axis.

3. What actually happens: If $\vec{B}$ is horizontal, the torque is horizontal and tries to tilt the loop so that $\vec{m}$ aligns with $\vec{B}$. But the loop is on a smooth horizontal plane — it can’t tilt because the plane prevents vertical motion. So the loop simply stays put, with no rotation at all.

Conclusion for (a): No, a uniform magnetic field cannot make the loop turn about the vertical axis.

(b) Orientation of stable equilibrium and maximum flux

Concept and intuition:

When a current loop is free to turn in a uniform external field, it behaves like a magnetic dipole. The potential energy is $U = -\vec{m} \cdot \vec{B}_{\text{ext}}$. The system seeks the minimum energy, which occurs when $\vec{m}$ is parallel to $\vec{B}_{\text{ext}}$ (i.e., the loop’s plane is perpendicular to the field). In that orientation, the flux of the total field through the loop is maximum — this is a consequence of Lenz’s law and energy minimization.

Step-by-step reasoning:

1. Energy of a magnetic dipole in an external field:

   The potential energy is $U = -\vec{m} \cdot \vec{B}_{\text{ext}} = -m B_{\text{ext}} \cos\theta$, where $\theta$ is the angle between $\vec{m}$ and $\vec{B}_{\text{ext}}$. Minimum energy occurs at $\theta = 0$ (parallel alignment), and maximum energy at $\theta = \pi$ (antiparallel). So stable equilibrium is when $\vec{m}$ is parallel to $\vec{B}_{\text{ext}}$.

2. Orientation of the loop:

   Since $\vec{m}$ is perpendicular to the plane of the loop (right-hand rule), $\vec{m} \parallel \vec{B}_{\text{ext}}$ means the plane of the loop is perpendicular to $\vec{B}_{\text{ext}}$. That is the stable orientation.

3. Flux of the total field:

   The total magnetic field at any point is $\vec{B}_{\text{total}} = \vec{B}_{\text{ext}} + \vec{B}_{\text{loop}}$, where $\vec{B}_{\text{loop}}$ is the field produced by the loop itself. The flux of $\vec{B}_{\text{total}}$ through the loop is:

The second term is the self-flux, which is $L I$ (where $L$ is the self-inductance), a constant for a given current. So maximizing $\Phi$ is equivalent to maximizing the external flux $\Phi_{\text{ext}} = \int \vec{B}_{\text{ext}} \cdot d\vec{A}$.

4. External flux: $\Phi_{\text{ext}} = B_{\text{ext}} A \cos\theta$, where $\theta$ is the angle between $\vec{B}_{\text{ext}}$ and the area vector $\vec{A}$ (which is parallel to $\vec{m}$). This is maximum when $\cos\theta = 1$, i.e., $\vec{A} \parallel \vec{B}_{\text{ext}}$, which is exactly the stable equilibrium orientation. So the total flux is maximum in that orientation.

Method: Energy Minimisation Principle for Magnetic Dipoles

A current-carrying loop behaves like a magnetic dipole with magnetic moment

$\vec{m} = I A \hat{n}$, where $I$ is current, $A$ is area, and $\hat{n}$ is the unit normal (right-hand rule).

In a uniform external magnetic field $\vec{B}_{\text{ext}}$, the potential energy is:

The system tends to minimise this energy to reach stable equilibrium.

(a) Can a uniform magnetic field make the loop turn about the vertical axis?

Answer: No.

Reason:

• The torque on a magnetic dipole in a uniform field is $\vec{\tau} = \vec{m} \times \vec{B}_{\text{ext}}$.
• For the loop to turn about the vertical axis, the torque must have a vertical component.
• But $\vec{m}$ is perpendicular to the plane of the loop. If the loop lies flat on a horizontal plane, $\vec{m}$ is vertical.
• A uniform $\vec{B}_{\text{ext}}$ can only produce a torque perpendicular to both $\vec{m}$ and $\vec{B}_{\text{ext}}$.
• Since $\vec{m}$ is vertical, any torque will be horizontal — it can tilt the loop, but cannot spin it about the vertical axis.

Key insight: A uniform field cannot produce a torque parallel to $\vec{m}$.

(b) Orientation of stable equilibrium and maximum flux

Step 1: Energy minimisation

$U = - m B_{\text{ext}} \cos\theta$, where $\theta$ is the angle between $\vec{m}$ and $\vec{B}_{\text{ext}}$.

• Minimum energy occurs when $\cos\theta = +1$, i.e., $\theta = 0$.
• Stable equilibrium orientation: $\vec{m}$ is parallel to $\vec{B}_{\text{ext}}$.

Step 2: Flux of total field

Total field: $\vec{B}_{\text{total}} = \vec{B}_{\text{ext}} + \vec{B}_{\text{loop}}$.

Flux through the loop: $\Phi = \int \vec{B}_{\text{total}} \cdot d\vec{A}$.

• When $\vec{m} \parallel \vec{B}_{\text{ext}}$, the loop’s own field adds to the external field inside the loop (both point in the same direction).
• This maximises the total flux through the loop.
• In any other orientation, the loop’s field partially opposes the external field, reducing flux. …

Here are the common mistakes students make on this Magnetic Force Balance question, along with how to avoid each.

(a) Loop turning about a vertical axis on a smooth horizontal plane

Common Mistake 1: Assuming any uniform horizontal field will cause rotation.

Students often think that because a current loop experiences a torque in a magnetic field, it will always rotate. They forget to check the direction of the torque vector.

• How to avoid: Recall the torque on a current loop: $\vec{\tau} = \vec{m} \times \vec{B}$, where $\vec{m}$ is the magnetic moment (perpendicular to the loop plane, given by the right-hand rule). For the loop to turn about a vertical axis, the torque must be vertical.
• The key insight: If the magnetic field is horizontal and uniform, the torque $\vec{\tau}$ is perpendicular to both $\vec{m}$ (vertical) and $\vec{B}$ (horizontal). This torque is horizontal, not vertical. A horizontal torque tries to tilt the loop, not spin it around a vertical axis. Since the loop is on a smooth horizontal plane, the plane prevents tilting. Therefore, no rotation occurs.

Common Mistake 2: Confusing "turning about vertical axis" with "spinning in the plane."

Students imagine the loop spinning like a coin on a table.

• How to avoid: Draw the vectors. The magnetic moment $\vec{m}$ is perpendicular to the loop. For the loop to spin in the plane (about a vertical axis), $\vec{m}$ must rotate in the horizontal plane. This requires a torque that is vertical (to change the horizontal direction of $\vec{m}$). A uniform horizontal $\vec{B}$ gives a horizontal torque, not a vertical one. A uniform vertical $\vec{B}$ gives $\vec{\tau} = \vec{m} \times \vec{B} = 0$ (since $\vec{m}$ and $\vec{B}$ are parallel). So, no uniform field can cause this rotation.

Final Answer for (a): No. A uniform magnetic field cannot make the loop turn about a vertical axis because the required torque direction is not achievable with a uniform field.

(b) Orientation of stable equilibrium and flux maximization

Common Mistake 3: Stating the stable orientation is "any orientation" or "perpendicular to the field."

Students often confuse the torque equation. They remember torque is zero when $\vec{m}$ is parallel or anti-parallel to $\vec{B}$, but forget which is stable.

• How to avoid: Use the potential energy formula: $U = -\vec{m} \cdot \vec{B} = -mB\cos\theta$.
  • Stable equilibrium is at minimum energy.
  • $U$ is minimum when $\cos\theta = +1$, i.e., $\theta = 0^\circ$.
  • This means $\vec{m}$ is parallel to $\vec{B}$. The loop plane is perpendicular to the field.
  • Unstable equilibrium is when $\vec{m}$ is anti-parallel ($\theta = 180^\circ$, maximum energy).

Common Mistake 4: Thinking "flux of total field" means just the external field flux.

Students calculate flux of only the external field through the loop, missing the loop's own field contribution.

• How to avoid: The question explicitly says "flux of the total field (external field + field produced by the loop)." This is a subtle but crucial point.
  • The flux of the loop's own field through itself is always positive (self-inductance, $L I$).
  • The flux of the external field through the loop is $\Phi_{ext} = \vec{B}_{ext} \cdot \vec{A} = BA\cos\theta$.
  • Total flux $\Phi_{total} = \Phi_{ext} + LI$.
  • In stable equilibrium ($\theta=0$), $\Phi_{ext}$ is maximum ($+BA$). Since $LI$ is constant, the total flux is maximum.
  • In unstable equilibrium ($\theta=180^\circ$), $\Phi_{ext}$ is minimum ($-BA$), so total flux is minimum.

Final Answer for (b): Stable equilibrium is when the plane of the loop is perpendicular to the external field, with the magnetic moment parallel to the field. In this orientation, the flux of the total field is maximum because the external flux is at its positive maximum and adds to the constant self-flux.

(c) Flexible irregular loop becoming circular

Common Mistake 5: Thinking the loop changes shape to "align with the field" or "maximize area." …