Question

A solenoid is made from a copper wire that is 15 m long and has a radius of 1.6 mm. The resistivity of copper is 1.68 x 10-8 Ωm. To form the solenoid, the copper wire is tightly wrapped around a tube that is 0.35 m long and has a diameter of 2.5 cm. The solenoid is connected to a power source which supplies an emf of 2.5 V. Find the magnitude of the magnetic field through the center of this solenoid

Answer 1

The magnitude of the magnetic field through the center of this solenoid is 0.0568 T.

Step-by-step explanation:

Given that,

Length = 15 m

Radius r₁ = 1.6 mm

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

Radius
`r_(2) =(2.5)/(2)=1.25\ cm`

emf = 2.5 V

Length of tube = 0.35 m

We need to calculate the area of cross section

`A = \pi r_(1)^2`

`A=\pi*(1.6*10^(-3))^2`

`A=0.000008042\ m^2`

`A=8.038*10^(-6)\ m^2`

We need to calculate the resistance

Using formula of resistivity

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

Put the value into the formula

`R=(1.68*10^(-8)*15)/(8.038*10^(-6))`

`R=0.03135\ \Omega`

We need to calculate the current

Using formula of current

`I=(\epsilon)/(R)`

Where,
`\epsilon` = emf

Put the value into the formula

`I=(2.5)/(0.03135)`

`I=79.745\ A`

We need to calculate the number of turns per unit length

Using formula of number of turns

`N=(l)/(2\pi r_(2))`

`N=(15)/(2\pi*1.25*10^(-2))`

`N=190.98`

`n=(N)/(l)`

`n=(190.98)/(0.35)`

`n=545.65\ turns/m`

We need to calculate the magnetic field

Using formula of magnetic field

`B=\mu_(0)nI`

`B=4\pi*10^(-7)*545.65*79.745`

`B=0.0568\ T`

Hence, The magnitude of the magnetic field through the center of this solenoid is 0.0568 T.