91影视

We have given that,

$R=0.2m$

$B=5脳{10}^{-3}T$

$D=1.02脳{10}^{-3}m$

$I_{max}=6A$

The solenoid is wounded in $N$turns, and due to the current in this wounded wire, a magnetic field in a uniform shape is produced at the center of the solenoid, where equation $29.17$gives the magnetic field everywhere inside the infinite solenoid, and gives the magnetic field along the central axis of an infinite solenoid by

$B=\frac{{渭}_{慰}NI}{l}\left(1\right)$

Where $N$denotes the number of turns,$L$ denotes the length, $I$denotes the current in the solenoid, and${渭}_{慰}$ denotes the permeability of free space and equals$4蟺脳{10}^{-7}T路m/A$.

By solving for$I$ we get

$I=\frac{Bl}{{渭}_{慰}N}\left(2\right)$

$18$gauge wire: The number of turns is equal to the length of the solenoid multiplied by the diameter of the thin layer, therefore for $18$gauge wire,

$N=\frac{0.20m}{1.02脳{10}^{-3}m}=196\text{turns}$

To calculate the current in the solenoid $18$gauge, insert the values for ${渭}_{慰},B,l\text{and}N$into equation $\left(2\right)$

$I_{18\text{guage}}=\frac{Bl}{{渭}_{慰}N}$

$$\begin{gathered} =\frac{(5脳{10}^{-3}T)(0.20m)}{(4蟺脳{10}^{-7}T路m/A)(196\text{turns)}} \\ =4.1A \end{gathered}$$

$26$gauge wire:

The number of turns for $26$gauge wire is calculated by multiplying the solenoid's length by the thin layer's diameter.

$N=\frac{0.20m}{0.41脳{10}^{-3}m}=488\text{turns}$

To calculate the current in the solenoid $26$gauge, insert the values for ${渭}_{慰},B,l\text{and}N$into equation $\left(2\right)$

$I_{26\text{guage}}=\frac{Bl}{{渭}_{慰}N}$

$$\begin{gathered} =\frac{(5脳{10}^{-3}T)(0.20m)}{(4蟺脳{10}^{-7}T路m/A)(488\text{turns)}} \\ =1.6A \end{gathered}$$

If we use $18$gauge wire, we want to pass $4.1A$, and if we use $26$gauge, we want to pass $1.6A$. Gauge $18$can carry up to $6A$of current, but gauge $26$can only carry $1A$. So, for a current of $4.1A$, the appropriate wire is $18$gauge.