Question

Marvin uses a long copper wire with resistivity 1.68 x 10^-8 Ω⋅m and diameter 1.00 x 10^−3​​ m to create a solenoid that has a 3.00 cm radius.
Part A: Calculate the total resistance if Marvin uses 11.3 meters of wire to create the solenoid.

Part B: Calculate how many turns are in the solenoid.

Part C: A 3.00 Volt potential difference is applied across the wires of the solenoid. Calculate the current that passes through the solenoid.

Part D: If the coils in the solenoid are separated so that they are evenly spaced across a total length of 0.1 m, what is the magnitude of the magnetic field along the central axis inside of the solenoid?

Answer 1

Answer:

Explanation:

resistivity of copper, ρ = 1.68 x 10^-8 ohm - m

diameter = 1 x 10^-3 m

radius of wire, r = 0.5 x 10^-3 m

Radius of solenoid, r' = 3 cm

(A)

Length of wire, l = 11.3 m

Let the resistance of wire is R.

$R= \frac{\rho\times l}{A}$

where, A is the crossection of wire

A = 3.14 x 0.5 x 10^-3 x 0.5 x 10^-3 = 7.85 x 10^-7 m²

$R= \frac{1.68\times 10^{-8}\times 11.3}{7.85\times 10^{-7}}$

R = 0.24 ohm

(B)

Let the number of turns is N.

Length of wire = N x circumference of one turn

11.3 = N x 2 x 3.14 x 0.03

N = 60

(C)

V = 3 V

R = 0.24 ohm

V = i x R

3 = i x 0.24

i = 12.5 A

(D) number of turns per unit length, n = N / 0.1 = 60 / 0.1 = 600

The magnetic field is given by

$B = \mu _{0}ni$

$B = 4 \times 3.14 \times 10^{-7}\times 600\times 12.5$

B = 9.42 x 10^-3 Tesla

Answer 2

Answer with Explanation:

We are given that

Resistivity of copper wire=$\rh0=1.68\times 10^{-8}\Omega m$

Diameter=d=$1.00\times 10^{-3} m$

Radius of copper wire=$r=\frac{d}{2}=\frac{1}{2}\times 10^{-3} m$

Radius of solenoid=r'$=3 cm=3\times 10^{-2} m$

1 m=100 cm

a.Length of wire=l=11.3 m

Area of wire=A=$\pi r^2$

Where $\pi=3.14$

A=$3.14\times (\frac{1}{2}\times 10^{-3})^2$

Resistance, R=$\rho \frac{l}{A}$

Using the formula

$R=1.68\times 10^{-8}\times\frac{11.3}{3.14\times (\frac{1}{2}\times 10^{-3})^2}$

$R=0.24\Omega$

B.Length of solenoid=$2\pi r'=2\times 3.14\times 3\times 10^{-2}=0.188$ m

Number of turns=$n_0=\frac{l}{2\pi r'}=\frac{11.3}{0.188}$

$n_0$=60

C.Potential difference,V=3 V

Current,I=$\frac{V}{R}$

I=$\frac{3}{0.24}=12.5 A$

D.Total length =0.1 m

Number of turns per unit length,n=$\frac{60}{0.1}=600$

Magnetic field along central axis inside of the solenoid,B=$\mu_0 nI$

$B=4\pi\times 10^{-7}\times 12.5\times 600=9.42\times 10^{-3} T$