Question

(8%) Problem 11: A solenoid is created by wrapping al - 25 m long wire around a hollow tube of diameter D=1.5 cm. The wire diameter is d = 0.6 mm. The solenoid wire is then connected to a power supply so that a current of 1 = 4.5A flows through the wire. Randomized Variables L = 25 m D=1.5 cm d 0.6 mm T= 45 A 20% Part (a) Write an expression for the number of turns in the solenoid. You do not need to take into account the diameter of the wire in this calculation NE Grade Summary Deductions 09 Potential 10090 7 8 9 Submissions a d 4 6 Attempts reming D b 2 3 per attempt detailed view k 1 0 m P t VO 7 5 1 1 Submit Hint Iep His deduction per hit Hints rema Feedback deduction per feedback 10 Part (b) Calculate the number of tums, in the solenoid 20 Part (0) Write an expression for the length of the solenoid (L) in terms of the diameter of the hollow tube the length of the wure and the diameter of the wired. Assume it is constructed by using only 1 layer of loops (note that most solenoids are actually constructed with many layers, to maximize the magnetic field density) 20 Part (d) Calculate the length of the solenoid (2) in meters 20. Parte Calculate the magnitude of the magnetic field at the center of the solenoid in tesla

Answer 1

Given:

L = length of wire = 25 m

D = 1.5 cm = 0.015 m

d = diameter of wire = 0.6 mm = 0.0006 m

I = 4.5 A

a) Let the number of turns is N.

Then, N*circurmference of loop formed = Length of wire

$\Rightarrow N*2\pi*\frac{D}{2}=L$

$\Rightarrow N=\frac{L}{\pi D}$.

Therefore, the expression of number of turns is $N=\frac{L}{\pi D}$ [answer]

b) Putting the values in the expression of number of turns in part a, we get the value of number of turns in the solenoid.

$N=\frac{L}{\pi D} = \frac{25}{\pi*0.015}\approx530$[answer]

c) Let the length of solenoid is L2.

The thickness of the wire is 'd', and the number of turns is 'N'(later we use the expression of N in place of N that we have deduced in part a).

Therefore, the length of the solenoid is

[answer]

Ld L L2 = N*d= =*d= D 7 TD

d) By putting the values in the expression of length L2, we get the value of L2.

$L_{2}=\frac{Ld}{\pi D}= \frac{25*0.0006}{\pi*0.015}=0.32 m$ [answer]

e) The magnitude of magnetic field at the center of the solenoid is given by $B=\frac{\mu_{o}NI}{\text{length of solenoid}}$

After putting the values, $B= \frac{\mu_{o}NI}{L_{2}}=\frac{4\pi*10^{-7}*530*4.5}{0.32}=0.0094T$ [answer]