A magnet is allowed to fall towards a metal ring. During the fall Its accelertion is

Options

A. Equal to g
B. Greater than g
C. Less than g
D. Equal to the product of g and the radius of the ring

Answer

To solve the problem of a magnet falling towards a metal ring, we need to analyze the forces acting on the magnet and the effects of electromagnetic induction. Here’s a step-by-step solution:

### Step 1: Understanding the Setup
When a magnet is allowed to fall towards a metal ring, it moves through a magnetic field. The magnet has a north and south pole, and as it approaches the ring, the magnetic flux through the area enclosed by the ring changes.

**Hint:** Remember that a changing magnetic flux induces an electromotive force (EMF) in the conductor (the metal ring).

### Step 2: Applying Faraday's Law of Induction
According to Faraday's law, a change in magnetic flux through a circuit induces an electromotive force (EMF) in that circuit. As the magnet falls, the magnetic flux through the ring increases, which induces a current in the ring.

**Hint:** Consider how the direction of the induced current is determined by Lenz's law.

### Step 3: Lenz's Law
Lenz's law states that the direction of the induced current will be such that it opposes the change in magnetic flux. In this case, as the magnet falls and the flux increases downward, the induced current will create a magnetic field that opposes this increase. Therefore, the induced magnetic field will have its north pole facing upward, repelling the north pole of the falling magnet.

**Hint:** Think about how the interaction between the induced magnetic field and the falling magnet affects the motion of the magnet.

### Step 4: Analyzing Forces on the Magnet
The forces acting on the falling magnet include:
- The gravitational force (weight) acting downward, \( mg \).
- The magnetic repulsive force from the induced current in the ring acting upward.

Let \( F \) be the magnetic force acting upward due to the induced current. The net force acting on the magnet can be expressed as:
\[ F_{\text{net}} = mg - F \]

**Hint:** Remember that the net force affects the acceleration of the magnet.

### Step 5: Applying Newton's Second Law
According to Newton's second law, the net force is also equal to the mass of the magnet times its acceleration \( a \):
\[ F_{\text{net}} = ma \]

Setting the two expressions for net force equal gives:
\[ ma = mg - F \]
From this, we can solve for acceleration \( a \):
\[ a = g - \frac{F}{m} \]

**Hint:** Consider how the magnitude of the magnetic force \( F \) affects the acceleration of the magnet.

### Step 6: Conclusion on Acceleration
Since the magnetic force \( F \) is always positive and acts upward, it reduces the effective acceleration of the magnet. Therefore, the acceleration \( a \) of the magnet is less than \( g \):
\[ a < g \]

Thus, the final conclusion is that the acceleration of the magnet as it falls towards the metal ring is less than the acceleration due to gravity.

**Final Answer:** The acceleration of the magnet during its fall is less than \( g \).