Figure shows two circular rings of radii a & b (a gt b) joined together by wire of negligible resistance. If the arrangement is placed in a time varying magnetic field, `(dB)/(dt)=k` and if the resistance per unit length of wire is `lambda`, then induced current is

Options

A. `(k(a-b))/(2lambda)`
B. `(k(a+b))/(2lambda)`
C. `(k(a^(2)-b^(2)))/(2lambda(a-b))`
D. `(k(a^(2)+b^(2)))/(2lambda(a+b))`

Answer

To solve the problem, we need to determine the induced current in the arrangement of two circular rings placed in a time-varying magnetic field. Here’s a step-by-step solution:

### Step 1: Understand the Setup
We have two circular rings with radii \( a \) (larger) and \( b \) (smaller), connected by a wire of negligible resistance. The magnetic field is changing with time, represented by \( \frac{dB}{dt} = k \).

### Step 2: Calculate the Induced EMF for Each Ring
The induced electromotive force (EMF) in each ring can be calculated using Faraday's law of electromagnetic induction, which states that the induced EMF (\( E \)) in a closed loop is equal to the negative rate of change of magnetic flux through the loop.

1. **For the larger ring (radius \( a \))**:
\[ E_1 = \frac{d\Phi}{dt} = \frac{d}{dt}(B \cdot A) = \frac{dB}{dt} \cdot \pi a^2 \]
Substituting \( \frac{dB}{dt} = k \):
\[ E_1 = k \cdot \pi a^2 \]

2. **For the smaller ring (radius \( b \))**:
\[ E_2 = \frac{d\Phi}{dt} = \frac{d}{dt}(B \cdot A) = \frac{dB}{dt} \cdot \pi b^2 \]
Substituting \( \frac{dB}{dt} = k \):
\[ E_2 = k \cdot \pi b^2 \]

### Step 3: Calculate the Net EMF
The net EMF (\( E_{\text{net}} \)) in the circuit formed by the two rings is the difference between the EMFs of the two rings:
\[ E_{\text{net}} = E_1 - E_2 = k \cdot \pi a^2 - k \cdot \pi b^2 \]
Factoring out \( k \cdot \pi \):
\[ E_{\text{net}} = k \cdot \pi (a^2 - b^2) \]

### Step 4: Calculate the Total Resistance
The total resistance \( R \) of the wire connecting the two rings can be calculated as follows:
- The circumference of the larger ring is \( 2\pi a \).
- The circumference of the smaller ring is \( 2\pi b \).
- The total length of the wire is \( 2\pi a + 2\pi b \).

The resistance per unit length is \( \lambda \), so the total resistance \( R \) is:
\[ R = \lambda \cdot (2\pi a + 2\pi b) = 2\pi \lambda (a + b) \]

### Step 5: Calculate the Induced Current
Using Ohm's law, the induced current \( I \) can be calculated as:
\[ I = \frac{E_{\text{net}}}{R} \]
Substituting the values we found:
\[ I = \frac{k \cdot \pi (a^2 - b^2)}{2\pi \lambda (a + b)} \]
Simplifying:
\[ I = \frac{k (a^2 - b^2)}{2\lambda (a + b)} \]

### Final Answer
The induced current is:
\[ I = \frac{k (a - b)(a + b)}{2\lambda (a + b)} = \frac{k (a - b)}{2\lambda} \]