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* 10^(-2) T`

Current,I=4.6 A

Diameter of wire,d=0.5 mm=
`0.5* 10^(-3) m`

Radius of wire,r=
`(d)/(2)=(0.5* 10^(-3))/(2)=0.25* 10^(-3) m`

`1mm=10^(-3) m`

Radius of solenoid,r'=1 cm=
`1* 10^(-2) m`

`1 cm=10^(-2) m`

Resistivity of copper,
`\rho=1.68* 10^(-8)\Omega m`

We know that

`R=(\rho l)/(A)`

Where
`A=\pi r^2`

Using the formula

`4.3=(1.68* 10^(-8)* l)/(\pi(0.25* 10^(-3))^2)`

`l=(4.3* \pi(0.25* 10^(-3))^2)/(1.68* 10^(-8))=50.23 m`

Number of turns of wire=
`(l)/(2\pi r')`

Number of turns of wire=
`(50.26)/(2\pi(1* 10^(-2))=800`

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

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

`3.7* 10^(-2)=4\pi* 10^(-7) n* 4.6`

`n=(3.7* 10^(-2))/(4* 3.14* 10^(-7)* 4.6)`

`n=6404 turns/m`

`n=(N)/(L)`

`L=(N)/(n)`

`L=(799)/(6404)`

`L=0.125 m=0.125* 100=12.5 cm`

Length of solenoid=12.5 cm

1m=100 cm