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NCERT Exemplar · Q10

Q.Consider a metal ring kept on top of a fixed solenoid (say on a cardboard). The centre of the ring coincides with the axis of the solenoid. If the current is suddenly switched on, the metal ring jumps up. Explain.

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When current is switched on in the solenoid, the magnetic flux through the metal ring increases rapidly. By Lenz’s law, the induced current in the ring opposes this change, creating a repulsive force that makes the ring jump upward.


Why does the ring jump? The core idea

This is a beautiful demonstration of Mutual Inductance and Lenz’s law in action. The solenoid and the metal ring are not connected electrically — but they are linked magnetically. When the solenoid’s current changes, the magnetic field it produces changes, and that changing field cuts through the ring.

The ring is a closed conducting loop. A changing magnetic flux through it induces an electromotive force (EMF), which drives a current in the ring. That induced current then experiences a force in the solenoid’s magnetic field. The direction of this force is such that it opposes the change that caused it — that’s Lenz’s law. Here, the change is the sudden increase of flux through the ring, so the ring tries to reduce that flux. It does so by moving away from the solenoid, where the field is weaker — hence it jumps up.


Step-by-step reasoning

1. The solenoid creates a magnetic field along its axis

When the switch is closed, current II flows through the solenoid. For a long solenoid, the magnetic field inside is nearly uniform and directed along the axis:

B=μ0nIB = \mu_0 n I

where nn is the number of turns per unit length. Outside the solenoid, the field is much weaker and diverges.

The metal ring is placed coaxially on top of the solenoid. Initially, the flux through the ring is zero (no current). As soon as the current is switched on, the field builds up, and magnetic field lines pass through the ring’s area.

Note

The ring is not a perfect conductor — it has some resistance. But the induced current is still large enough to produce a noticeable effect.

2. The changing flux induces an EMF in the ring

The magnetic flux Φ\Phi through the ring is:

Φ=∫B⋅dA\Phi = \int \mathbf{B} \cdot d\mathbf{A}

Since the ring is coaxial, the field is perpendicular to its plane, so Φ=BA\Phi = B A (approximately, if the ring is close to the solenoid’s end). As BB rises from 0 to its steady value, Φ\Phi changes. Faraday’s law gives the induced EMF:

E=−dΦdt\mathcal{E} = -\frac{d\Phi}{dt}

This EMF drives an induced current IringI_{\text{ring}} in the ring. The direction of IringI_{\text{ring}} is such that its own magnetic field opposes the increase in flux — that is, it tries to push back against the solenoid’s field.

3. The induced current experiences a force in the solenoid’s field

Now, the ring carrying current IringI_{\text{ring}} sits in the non-uniform magnetic field of the solenoid (especially near the end). A current-carrying loop in a magnetic field experiences a net force if the field is non-uniform. The force on a small element dld\mathbf{l} of the ring is:

dF=Iring dl×Bd\mathbf{F} = I_{\text{ring}} \, d\mathbf{l} \times \mathbf{B} …

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