You are working as an electrical technician. One day, out in CR the field, you need an inductor but cannot find one. Look- ing in your wire supply cabinet, you find a cardboard tube with single-conductor wire wrapped uniformly around it to form a solenoid. You carefully count the turns of wire and find that there are 580 turns. The diameter of the tube is 8.00 cm, and the length of the wire-wrapped portion is 36.0 cm. You pull out your calculator to determine: a. the inductance of the coil
b. the emf generated in it if the current in the wire increases at the rate of 4.00 A/s.

Explanation:

Given info: The Number of turns in the wire is 580 the diameter of  

tube is and the length of the tube up to which wire is wrapped is .  

Formula to calculate the inductance of the coil is,  

Here,  

is inductance of the coil.  

is the permittivity.  

is the number of turns.  

is the length up to which wire is wrapped.  

is the cross sectional area of the coil.  

The expression for the area is,  

Substitute in equation (1).  

Substitute 580 for  

Conclusion:  

Therefore, the inductance of the given single conductor wire is  

Given info: The rate of increasing current and the inductance of the coil .

The generated emf is,

Here,

is the generated emf.

Conclusion:

Therefore, the generated emf is .

b) phosphorus pentoxide

i think the answer is a and b

i wish you luck