It is desired to construct a solenoid that will have a resistance of 5.80 omegac

Question

It is desired to construct a solenoid that will have a resistance of 5.80 omegacap.gif (at 20C) and produce a magnetic field of 4.00 ? 10-2 T at its center when it carries a current of 4.80A. The solenoid is to be constructed from copper wire having a diameter of 0.500 mm. If the radius of the solenoid is to be 1.00 cm, determine the following. (The resistivity of copper at 20C is 1.7 multiply.gif 10-8 ? m.) (a) the number of turns of wire needed to build the solenoid turns (b) the length the solenoid should have cm

Explanation / Answer

Area of wire

A=pi*d2/4 =pi*(0.5*10^-3)^2/4

A=1.963*10^-7 m2

Resistance

R=pL/A

The required length of wire

L=RA/p =5.8*(1.963*10^-7)/(1.7*10^-8)

L=67 m

so Total Number of turns on the solenoid is

N=L/2pir =67/2pi*(0.01) =1067 turns

b)

Magnetic field of a solenoid is given by

B =uo*N*I/L

4*10^-2 =(4pi*10^-7)*1067*4.8/L

L=0.16 m=16.1 cm

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.