# A solenoid is designed to produce a magnetic field of 0.030 T at its center. It has radius 1.50 cm and length 50.0 cm, and the wire can carry a maximum current of 11.0 A. (a) What minimum number of turns per unit length must the solenoid have? (b) What total length of wire is required?

This question aims to find the number of turns in a solenoid for a specific configuration and the total length of the wire.

The question depends on the concept of the solenoid. A solenoid is a coil made with conducting wire like copper. When a current passes through it, it generates a magnetic flux density around it which depends on the magnetic constant, number of turns in the coil, current, and length of the solenoid. The equation for the magnetic flux of the solenoid is given as:

$B = \mu_0 \dfrac{ NI }{ l }$

$B = Magnetic\ Flux$

$\mu_0 = Magnetic\ Constant$

$I = Current$

$l = Length\ of\ the\ Solenoid$

The given information for this problem is:

$B = 0.030\ T$

$Radius\ of\ the\ Coil\ r = 1.50 cm$

$Length\ of\ the\ Coil\ l = 50.0 cm$

$Current\ through\ the\ Coil\ I = 11.0 A$

$Magnetic\ Constant\ \mu_0 = 4 \pi \times 10^{-7} T.m/A$

a) To find the total number of turns in the coil, we can use the solenoid formula. The formula is given as:

$B = \mu_0 \dfrac{ NI }{ l }$

Rearranging the formula to find the number of turns in the coil as:

$N = \dfrac{ Bl }{ \mu_0 I }$

Substituting the values, we get:

$N = \dfrac{ 0.030 \times 0.5 }{ 4 \pi \times 10^ {-7} \times 11 }$

$N = \dfrac{ 0.015 }{ 138.23 \times 10^ {-7}}$

$N = 1085\ turns$

b) To find the length of the wire of the solenoid, we can use the number of turns in the solenoid and multiply it with the length of one turn which is given by the formula of the circumference of the circle. We know the radius of the solenoid, so we can find the total length of the wire by taking the product of number of turns and circumference of each turn. The length of the wire is given as:

$L = N \times 2 \pi r$

$r = 1.50 cm$

$N = 1085 turns$

Substituting the values, we get:

$L = 1085 \times 2 \pi \times 0.015$

$L = 1085 \times 0.094$

$L = 102.3 m$

## Numerical Result

a) The total number of turns in the solenoid that generates a 0.030 T of magnetic flux with a length of 50 cm and 11 A current is calculated to be:

$N = 1085 turns$

b) The total length of the wire of the same solenoid is calculated to be:

$L = 102.3 m$

## Example

Find the number of turns in a solenoid with length of 30 cm and 5 A current. It generates a 0.01 T of magnetic flux.

$Magnetic\ Flux\ B = 0.01 T$

$Current\ I = 5 A$

$Length\ of\ the\ Solenoid\ l = 0.3 m$

$Magnetic\ Constant\ \mu_0 = 4 \pi \times 10^ {-7} T.m/A$

The formula for total number of turns in the solenoid is given as:

$N = \dfrac{ Bl }{ \mu_0 I }$

Substituting the values, we get:

N = 0.01^5 / [4piX10^(-7)] X 0.3

N = 132629 turns

The total turns of the solenoid are calculated to be 132629 turns.