Mass m of a material of density d and resistivity p is used to make a solenoid of length l. Total resistance of the wire is R. Self-inductance of the solenoid is

Options

A. `(mu_(0)mR)/(4pi rho dl)`
B. `(mu_(0)mR)/(pi rho dl)`
C. `(4mu_(0)mR)/(pi rho dl)`
D. `(mu_(0)mlR)/(4pi rho dl)`

Explanation

To find the self-inductance \( L \) of a solenoid made from a material with mass \( m \), density \( d \), resistivity \( \rho \), and length \( l \), we can follow these steps:

### Step 1: Relate mass, density, and volume
We know that the mass \( m \) of the material is related to its density \( d \) and volume \( V \) by the formula:
\[ m = d \cdot V \]
For a solenoid, the volume \( V \) can be expressed as:
\[ V = A \cdot l \]
where \( A \) is the cross-sectional area of the solenoid.

### Step 2: Express volume in terms of area and length
Substituting the expression for volume into the mass equation gives:
\[ m = d \cdot (A \cdot l) \]
From this, we can express the cross-sectional area \( A \):
\[ A = \frac{m}{d \cdot l} \]

### Step 3: Relate resistance to resistivity
The resistance \( R \) of the wire used to make the solenoid can be expressed in terms of its resistivity \( \rho \), length \( x \), and cross-sectional area \( A \):
\[ R = \frac{\rho \cdot x}{A} \]
Substituting the expression for \( A \) from Step 2:
\[ R = \frac{\rho \cdot x \cdot d \cdot l}{m} \]

### Step 4: Express length of the wire in terms of turns
For a solenoid, the total length of the wire \( x \) can be expressed in terms of the number of turns \( n \) and the radius \( r \):
\[ x = 2 \pi r n l \]

### Step 5: Substitute \( x \) into the resistance equation
Now substituting the expression for \( x \) into the resistance equation:
\[ R = \frac{\rho \cdot (2 \pi r n l) \cdot d \cdot l}{m} \]
This simplifies to:
\[ R = \frac{2 \pi \rho d n l^2 r}{m} \]

### Step 6: Find self-inductance
The self-inductance \( L \) of the solenoid can be expressed as:
\[ L = \mu_0 \cdot n^2 \cdot A \cdot l \]
Substituting \( A \) from Step 2:
\[ L = \mu_0 \cdot n^2 \cdot \left(\frac{m}{d \cdot l}\right) \cdot l \]
This simplifies to:
\[ L = \frac{\mu_0 m n^2}{d} \]

### Step 7: Substitute \( n \) in terms of \( R \)
From the resistance equation, we can express \( n \):
\[ n = \frac{m R}{2 \pi \rho d l^2} \]
Substituting this back into the self-inductance equation gives:
\[ L = \frac{\mu_0 m}{d} \cdot \left(\frac{m R}{2 \pi \rho d l^2}\right)^2 \]
After simplifying, we arrive at:
\[ L = \frac{\mu_0 m R}{4 \pi \rho d l} \]

### Final Answer
Thus, the self-inductance \( L \) of the solenoid is:
\[ L = \frac{\mu_0 m R}{4 \pi \rho d l} \]

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