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 determine the induced electric field at a distance \( r \) from the axis of two coaxial solenoids. Here's a step-by-step solution:

### Step 1: Define the currents in the solenoids
Let the current in the outer solenoid be \( I_{\text{outer}} = Kt \). Since the current in the inner solenoid is twice that of the outer one, we have:
\[ I_{\text{inner}} = 2I_{\text{outer}} = 2Kt \]

### Step 2: Determine the magnetic field produced by each solenoid
The magnetic field \( B \) inside a solenoid is given by:
\[ B = \mu_0 n I \]
where \( \mu_0 \) is the permeability of free space, \( n \) is the number of turns per unit length, and \( I \) is the current.

For the inner solenoid:
\[ B_{\text{inner}} = \mu_0 n (2Kt) = 2\mu_0 n Kt \]

For the outer solenoid:
\[ B_{\text{outer}} = \mu_0 n (Kt) = \mu_0 n Kt \]

### Step 3: Calculate the total magnetic field at a distance \( r \) (where \( R < r < 2R \))
In the region between the two solenoids (i.e., \( R < r < 2R \)), the magnetic field due to the inner solenoid will dominate since the outer solenoid's field is zero in that region. Thus, the total magnetic field \( B \) at distance \( r \) is:
\[ B = B_{\text{inner}} = 2\mu_0 n Kt \]

### Step 4: Calculate the magnetic flux through a circular loop of radius \( r \)
The magnetic flux \( \Phi \) through a circular loop of radius \( r \) is given by:
\[ \Phi = B \cdot A = B \cdot \pi r^2 \]
Substituting for \( B \):
\[ \Phi = (2\mu_0 n Kt) \cdot \pi r^2 = 2\mu_0 n Kt \pi r^2 \]

### Step 5: Determine the induced electric field using Faraday's Law
According to Faraday's Law of electromagnetic induction, the induced electromotive force (emf) is equal to the negative rate of change of magnetic flux:
\[ \mathcal{E} = -\frac{d\Phi}{dt} \]
Calculating the derivative:
\[ \frac{d\Phi}{dt} = 2\mu_0 n K \pi r^2 \]
Thus, the induced emf is:
\[ \mathcal{E} = -2\mu_0 n K \pi r^2 \]

### Step 6: Relate the induced emf to the electric field
The induced emf around a circular path of radius \( r \) is also given by:
\[ \mathcal{E} = E \cdot 2\pi r \]
Setting the two expressions for emf equal gives:
\[ E \cdot 2\pi r = -2\mu_0 n K \pi r^2 \]
Solving for \( E \):
\[ E = -\frac{2\mu_0 n K \pi r^2}{2\pi r} = -\frac{\mu_0 n K r}{1} \]
Thus, the induced electric field at a distance \( r \) from the axis is:
\[ E = -\frac{\mu_0 n K r}{1} \]

### Final Answer
The induced electric field at a distance \( r \) from the axis, where \( R < r < 2R \), is:
\[ E = -\frac{\mu_0 n K r}{1} \]