A long, straight vertical copper wire has a 0.350-mu T magnetic field 17.0 cm pe

Question

A long, straight vertical copper wire has a 0.350-mu T magnetic field 17.0 cm perpendicular from the wire. Calculate the current needed to create this magnetic field. This is similar to Example #6 (page 10) of Chapter 21 notes. _____ You wish to generate a magnetic field of 500. HT at the center of a 25.0-cm long air solenoid. If you pass 1.50 A through the wire, how many turns of wire will be required to generate this magnetic field which equals that of Jupiter's? This is similar to Example #7 (page 10) of Chapter 21 notes. _____

Explanation / Answer

Q5.

magnetic field due to a current carrying wire is given by

B=mu*I/(2*pi*d)

where mu=magnetic permeability

I=current in the wire

d=distance from the wire

given B=0.35*10^(-6) T

d=0.17 m

then I=2*pi*d*B/mu

=2*pi*0.17*0.35*10^(-6)/(4*pi*10^(-7))=0.2975 A

hence current required is 0.2975 A

Q6.

magnetic field at the center of air solenoid is given by

B=mu*N*I/L

where N=number of turns

L=length

given B=500 uT=500*10^(-6) T

I=1.5 A

L=0.25 m

then N=B*L/(mu*I)

=500*10^(-6)*0.25/(4*pi*10^(-7)*1.5)=66.314

taking the nearest largest integer,

number of turns required is 67.

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