A thin wire ring of radius a and resitance `r` is located inside a long solenoid is equal to `l`, its cross-sectional radius , to `b`. At a certain moment the solenoid was connected to a source of a constant voltage `V`. The total resistance of the circuit is equal to `R`. Assuming the the radial force acting per unit length of the ring.

Answer

To solve the problem of finding the radial force acting per unit length on a thin wire ring located inside a long solenoid, we will follow these steps:

### Step 1: Understanding the Setup
We have a thin wire ring of radius \( a \) and resistance \( r \) inside a long solenoid of length \( l \) and cross-sectional radius \( b \). The solenoid is connected to a constant voltage source \( V \).

### Step 2: Magnetic Field Inside the Solenoid
The magnetic field \( B \) inside a long 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 of the solenoid, and \( I \) is the current flowing through the solenoid.

### Step 3: Induced EMF in the Ring
When the solenoid is connected to the voltage source, a current \( I \) flows through it, creating a changing magnetic field. The magnetic flux \( \Phi \) through the ring is given by:
\[ \Phi = B \cdot A = B \cdot \pi a^2 = \mu_0 n I \cdot \pi a^2 \]
The induced EMF \( \mathcal{E} \) in the ring can be calculated using Faraday's law of electromagnetic induction:
\[ \mathcal{E} = -\frac{d\Phi}{dt} \]

### Step 4: Current in the Ring
The current \( I' \) in the ring due to the induced EMF is given by Ohm's law:
\[ I' = \frac{\mathcal{E}}{r} \]

### Step 5: Force on the Ring
The force \( F \) acting on a current-carrying conductor in a magnetic field is given by:
\[ F = I' L B \]
where \( L \) is the length of the conductor (in this case, the circumference of the ring, \( 2\pi a \)). The force per unit length \( f \) is given by:
\[ f = \frac{F}{L} = I' B \]

### Step 6: Substitute Values
Substituting the expression for \( I' \) into the force per unit length:
\[ f = \left(\frac{\mathcal{E}}{r}\right) B \]
Now substituting \( \mathcal{E} = -\frac{d\Phi}{dt} \) and \( B = \mu_0 n I \):
\[ f = \left(\frac{-\frac{d}{dt}(\mu_0 n I \cdot \pi a^2)}{r}\right) B \]

### Step 7: Final Expression
After substituting and simplifying, we can express the radial force per unit length acting on the ring as:
\[ f = \frac{\mu_0 n^2 \pi a^2 V^2}{r l^2} e^{-\frac{R}{L}t} \left(1 - e^{-\frac{R}{L}t}\right) \]

### Conclusion
Thus, the radial force acting per unit length on the ring can be expressed in terms of the parameters of the solenoid and the ring.

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