A copper ring having a cut such as not to form a complete loop is held horizontally and a bar magnet is dropped through the ring. The acceleration of the falling magnet is \dots
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If the ring were a complete, closed solid loop, induced current would flow, creating an opposing "North" pole that acts like a parachute, slowing the magnet down (making $a < g$). The "cut" disables this entire magnetic braking effect.
Step 1: Understanding the Question:
A bar magnet is dropped through a copper ring. We need to determine its acceleration. Crucially, the problem explicitly notes that the copper ring "has a cut" preventing it from forming a complete, unbroken loop. Step 2: Key Formula or Approach:
1. Faraday's Law states that a changing magnetic flux will induce an Electromotive Force (EMF) in the ring.
2. Ohm's Law ($I = E/R$) states that current will only flow if there is a complete, closed electrical circuit.
3. Lenz's Law states that an induced current will create a magnetic field that opposes the motion of the magnet causing it. Step 3: Detailed Explanation:
As the magnet falls toward the ring, the magnetic flux passing through the area of the ring changes rapidly.
According to Faraday's law, an EMF is successfully induced across the copper material.
However, because the ring contains a physical cut, it represents an open circuit with infinite resistance ($R = \infty$).
Therefore, absolutely no induced current can flow around the ring ($I = 0$).
Because there is no current flowing, the ring acts like a dead piece of metal. It generates no opposing magnetic field of its own.
Since there is no upward repulsive magnetic force to counteract gravity, the only force acting on the falling magnet is its own weight ($F = mg$).
Thus, it falls freely under the pure influence of gravity.
Acceleration $a = g$. Step 4: Final Answer:
The acceleration is exactly $g$, matching option (a).
