A conducting ring of radius 2 metre is placed in an uniform magnetic field B of 0.01 Tesla oscillating with frequency 200 Hz with its plane at right angles to B . What will be the induced electric field ?

Options

A. 4 volts/m
B. 6 volts /m
C. 10 volts/m
D. 8 volts/m

Explanation

To solve the problem of finding the induced electric field in a conducting ring placed in a uniform magnetic field, we can follow these steps:

### Step 1: Understand the Problem
We have a conducting ring with a radius \( R = 2 \) meters placed in a uniform magnetic field \( B = 0.01 \) Tesla. The magnetic field is oscillating with a frequency \( f = 200 \) Hz, and the plane of the ring is perpendicular to the magnetic field.

### Step 2: Calculate the Area of the Ring
The area \( A \) of the ring can be calculated using the formula for the area of a circle:
\[ A = \pi R^2 \]
Substituting the value of \( R \):
\[ A = \pi (2)^2 = 4\pi \, \text{m}^2 \]

### Step 3: Calculate the Time Period of Oscillation
The time period \( T \) of the oscillation can be calculated using the frequency:
\[ T = \frac{1}{f} = \frac{1}{200} \, \text{s} = 0.005 \, \text{s} \]

### Step 4: Calculate the Change in Magnetic Flux
The magnetic flux \( \Phi \) through the ring is given by:
\[ \Phi = B \cdot A \]
Since the magnetic field is oscillating, we need to consider the change in flux over time. The maximum change in flux occurs when the angle \( \theta \) changes from \( 0 \) to \( 90 \) degrees.

At \( t = 0 \):
\[ \Phi_0 = B \cdot A = 0.01 \cdot 4\pi = 0.04\pi \, \text{Wb} \]
At \( t = \frac{T}{4} \) (where \( \theta = 90^\circ \)):
\[ \Phi = 0 \, \text{Wb} \]
Thus, the change in flux \( \Delta \Phi \) is:
\[ \Delta \Phi = \Phi - \Phi_0 = 0 - 0.04\pi = -0.04\pi \, \text{Wb} \]

### Step 5: Calculate the Induced EMF
According to Faraday's law of electromagnetic induction, the induced EMF \( \mathcal{E} \) is given by:
\[ \mathcal{E} = -\frac{d\Phi}{dt} \]
The average rate of change of flux over the time period \( T/4 \) is:
\[ \mathcal{E} = -\frac{\Delta \Phi}{T/4} = -\frac{-0.04\pi}{0.005/4} = \frac{0.04\pi \cdot 4}{0.005} = \frac{0.16\pi}{0.005} = 32\pi \, \text{V} \]

### Step 6: Relate Induced EMF to Electric Field
The induced electric field \( E \) around the ring can be found using the relationship between EMF and electric field:
\[ \mathcal{E} = E \cdot (2\pi R) \]
Substituting for \( R = 2 \):
\[ 32\pi = E \cdot (2\pi \cdot 2) \]
Simplifying gives:
\[ 32\pi = E \cdot 4\pi \]
Thus,
\[ E = \frac{32\pi}{4\pi} = 8 \, \text{V/m} \]

### Final Answer
The induced electric field is:
\[ \boxed{8 \, \text{V/m}} \]