A long solenoid of cross-sectional radius `a` has a thin insulates wiere ring tightly put on its winding, one half of the ring has the resistance `eta` times that of the other half. The magneticv induction produced by the solenoid varies with the time as `B = bt`, where `b` is a constant. Find the magnitude of the electric field strength in the ring.

Options

A. `(9)/(11)Rb`
B. `(9)/(22)Rb`
C. 9Rb
D. Rb

Answer

To solve the problem, we need to find the magnitude of the electric field strength in the wire ring surrounding the solenoid. The solenoid produces a changing magnetic field, which induces an electromotive force (emf) in the ring according to Faraday's law of electromagnetic induction.

### Step-by-Step Solution:

1. **Identify the Magnetic Field**:
The magnetic induction (B) produced by the solenoid varies with time as \( B = bt \), where \( b \) is a constant.

2. **Apply Faraday's Law**:
According to Faraday's law, the induced emf (\( \mathcal{E} \)) in the ring is given by:
\[ \mathcal{E} = -\frac{d\Phi_B}{dt} \]
where \( \Phi_B \) is the magnetic flux through the ring.

3. **Calculate the Magnetic Flux**:
The magnetic flux \( \Phi_B \) through the ring is given by:
\[ \Phi_B = B \cdot A = B \cdot \pi a^2 \]
where \( A \) is the area of the ring, and \( a \) is the radius of the cross-section of the solenoid.

4. **Differentiate the Magnetic Flux**:
Since \( B = bt \), we have:
\[ \Phi_B = (bt) \cdot \pi a^2 \]
Now, differentiate \( \Phi_B \) with respect to time \( t \):
\[ \frac{d\Phi_B}{dt} = \pi a^2 b \]

5. **Calculate the Induced EMF**:
Substitute the expression for \( \frac{d\Phi_B}{dt} \) into Faraday's law:
\[ \mathcal{E} = -\frac{d\Phi_B}{dt} = -\pi a^2 b \]

6. **Determine the Current in the Ring**:
The ring has two halves with different resistances. Let the resistance of one half be \( R_1 = \eta R \) and the other half \( R_2 = R \). The total resistance \( R_{total} \) is:
\[ R_{total} = R_1 + R_2 = \eta R + R = (\eta + 1) R \]

7. **Calculate the Current**:
Using Ohm's law, the current \( I \) in the ring can be calculated as:
\[ I = \frac{\mathcal{E}}{R_{total}} = \frac{-\pi a^2 b}{(\eta + 1) R} \]

8. **Find the Electric Field Strength**:
The electric field \( E \) in the ring can be found using the relationship between the electric field and the current:
\[ E = \frac{\mathcal{E}}{L} \]
where \( L \) is the circumference of the ring, \( L = 2\pi a \). Thus:
\[ E = \frac{-\pi a^2 b}{(\eta + 1) R \cdot 2\pi a} = \frac{-a b}{2(\eta + 1) R} \]

### Final Result:
The magnitude of the electric field strength in the ring is:
\[ E = \frac{ab}{2(\eta + 1) R} \]