Two coaxial solenoids having same number of turns per unit length n, and radius R and `2R` respectively have no current initially. Current in both grows uniformly such that current in the inner one is twice that of the outer one, at any instant of time in the same direction. Current in outer one at time `t` is `Kt` then induced electric field at a distance r from axis is `(R lt r lt 2R)`.

Options

A. Not possible
B. `(mu_(0)nK)/(2r)(2R^(2) +r^(2))`
C. `(2mu_(0)nKR^(2))/(r)`
D. `(2mu_(0)nK)/(r)(2R^(2)-r^(2))`

Explanation

To solve the problem, we need to find the induced electric field at a distance \( r \) from the axis of two coaxial solenoids. Let's break down the solution step by step.

### Step 1: Understand the current in the solenoids
- Let the current in the outer solenoid (radius \( 2R \)) be \( I_2 = Kt \).
- Since the current in the inner solenoid (radius \( R \)) is twice that of the outer one, we have \( I_1 = 2I_2 = 2Kt \).

### Step 2: Calculate the magnetic field in the solenoids
- The magnetic field inside a solenoid is given by the formula:
\[ B = \mu_0 n I \]
- For the inner solenoid (radius \( R \)):
\[ B_1 = \mu_0 n I_1 = \mu_0 n (2Kt) = 2\mu_0 n Kt \]
- For the outer solenoid (radius \( 2R \)):
\[ B_2 = \mu_0 n I_2 = \mu_0 n Kt \]

### Step 3: Calculate the magnetic flux through the area between the solenoids
- The area of the inner solenoid is:
\[ A_1 = \pi R^2 \]
- The area of the outer solenoid (considering the region between \( R \) and \( 2R \)) is:
\[ A_2 = \pi (2R)^2 - \pi R^2 = \pi (4R^2 - R^2) = 3\pi R^2 \]
- The total magnetic flux \( \Phi \) through the area between \( R \) and \( 2R \) is the sum of the flux due to both solenoids:
\[ \Phi = \Phi_1 + \Phi_2 = B_1 A_1 + B_2 A_2 \]
\[ \Phi = (2\mu_0 n Kt)(\pi R^2) + (\mu_0 n Kt)(3\pi R^2) \]
\[ \Phi = \mu_0 n Kt \pi R^2 (2 + 3) = 5\mu_0 n Kt \pi R^2 \]

### Step 4: Calculate the induced emf
- The induced emf \( \mathcal{E} \) is given by Faraday's law of electromagnetic induction:
\[ \mathcal{E} = -\frac{d\Phi}{dt} \]
- Differentiate the total flux with respect to time \( t \):
\[ \mathcal{E} = -\frac{d}{dt}(5\mu_0 n Kt \pi R^2) = -5\mu_0 n K \pi R^2 \]

### Step 5: Relate induced emf to the electric field
- The induced emf around a closed loop is also given by:
\[ \mathcal{E} = E \cdot L \]
where \( L \) is the circumference of the loop at radius \( r \):
\[ L = 2\pi r \]
- Therefore, we have:
\[ E \cdot (2\pi r) = 5\mu_0 n K \pi R^2 \]
- Solving for \( E \):
\[ E = \frac{5\mu_0 n K R^2}{2r} \]

### Final Answer
The induced electric field \( E \) at a distance \( r \) from the axis, where \( R < r < 2R \), is given by:
\[ E = \frac{5\mu_0 n K R^2}{2r} \]