A neutral metallic ring is placed in a circular symmetrical uniform magnetic field with its plane perpendicular to the field.If the magnitude of field starts increasing with time, then:

Options

A. the ring starts translating
B. the ring starts rotating about its axis
C. the ring slightly contracts
D. the ring starts rotating about a diameter

Answer

To solve the problem, we need to analyze the behavior of a neutral metallic ring placed in a circular symmetrical uniform magnetic field, particularly when the magnetic field's magnitude starts increasing with time.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have a neutral metallic ring in a uniform magnetic field.
- The plane of the ring is perpendicular to the magnetic field.

2. **Magnetic Field Change**:
- The magnetic field \( B \) is increasing with time, which implies \( \frac{dB}{dt} > 0 \).

3. **Induced EMF and Current**:
- According to Faraday's law of electromagnetic induction, a changing magnetic field through a closed loop induces an electromotive force (EMF) in that loop.
- The induced EMF (\( \mathcal{E} \)) can be calculated using the formula:
\[ \mathcal{E} = -\frac{d\Phi_B}{dt} \]
- Where \( \Phi_B \) is the magnetic flux given by \( \Phi_B = B \cdot A \) (with \( A \) being the area of the ring).

4. **Direction of Induced Current**:
- The direction of the induced current can be determined using Lenz's Law, which states that the induced current will flow in a direction that opposes the change in magnetic flux.
- Since the magnetic field is increasing, the induced current will flow in a direction that creates a magnetic field opposing the increase.

5. **Force on the Ring**:
- The induced current creates a magnetic moment \( \mathbf{m} \) in the ring.
- The force on a current-carrying conductor in a magnetic field is given by \( \mathbf{F} = I \mathbf{L} \times \mathbf{B} \).
- For a closed loop (the ring), the net force is zero because the lengths of the segments cancel out.

6. **Torque on the Ring**:
- The torque \( \tau \) on the ring due to the magnetic field is given by:
\[ \tau = \mathbf{m} \times \mathbf{B} \]
- Since the plane of the ring is perpendicular to the magnetic field, the angle \( \theta \) between \( \mathbf{m} \) and \( \mathbf{B} \) is zero, leading to:
\[ \tau = mB \sin(0) = 0 \]
- Therefore, there is no torque acting on the ring, meaning it cannot rotate.

7. **Contraction of the Ring**:
- The ring will try to oppose the increase in magnetic flux through it.
- To reduce the magnetic flux, the ring can contract, which decreases its area \( A \) and thus the magnetic flux \( \Phi_B \).

8. **Conclusion**:
- The ring does not translate or rotate, but it may contract in response to the increasing magnetic field.

### Final Answer:
The correct option is that the ring will **contract**.