Question

We want to construct a solenoid with a resistance of 4.30 Ω and generate a magnetic field of 3.70 × 10−2 T at its center when applying 4.60 A of electrical current. We want to use copper wire with a diameter of 0.500 mm. If we need the solenoid's radius to be 1.00 cm, how many turns of wire will be need? 2 78 256 790 Then, what would be the required length of the solenoid? 12.3 cm

Answer 1

Answer with Explanation:

We are given that

Resistance of solenoid,R=4.3 ohm

Magnetic field,B=$3.7\times 10^{-2} T$

Current,I=4.6 A

Diameter of wire,d=0.5 mm=$0.5\times 10^{-3} m$

Radius of wire,r=$\frac{d}{2}=\frac{0.5\times 10^{-3}}{2}=0.25\times 10^{-3} m$

$1mm=10^{-3} m$

Radius of solenoid,r'=1 cm=$1\times 10^{-2} m$

$1 cm=10^{-2} m$

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

We know that

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

Where $A=\pi r^2$

Using the formula

$4.3=\frac{1.68\times 10^{-8}\times l}{\pi(0.25\times 10^{-3})^2}$

$l=\frac{4.3\times \pi(0.25\times 10^{-3})^2}{1.68\times 10^{-8}}=50.23 m$

Number of turns of wire=$\frac{l}{2\pi r'}$

Number of turns of wire=$\frac{50.26}{2\pi(1\times 10^{-2}}=800$

Hence, the number of turns of the  solenoid,N=799

Magnetic field in solenoid,B=$\mu_0 nI$

$3.7\times 10^{-2}=4\pi\times 10^{-7} n\times 4.6$

$n=\frac{3.7\times 10^{-2}}{4\times 3.14\times 10^{-7}\times 4.6}$

$n=6404 turns/m$

$n=\frac{N}{L}$

$L=\frac{N}{n}$

$L=\frac{799}{6404}$

$L=0.125 m=0.125\times 100=12.5 cm$

Length of solenoid=12.5 cm

1m=100 cm