Chapter 29, Problem 055 A long solenold with 13.5 tums/cm and a radlus of 6.90 c

Question

Chapter 29, Problem 055 A long solenold with 13.5 tums/cm and a radlus of 6.90 cm caries a current of 20.7 mA A curent of 8.52 A exists n a stra ght conductor located along the central axls of the solenold. (a) At what radlal distance from the axls In centimeters wll the direction of the resulting magnetic feld be at 55.59 to the axial dlrection? (b) What is the magnitude of the magnetic fleld there? (a) Number (b) Number Click if you would like to Show Work for this question: Open Show Work Units Units LINK TO TEXT LINK TO SAMPLE PROBLEM Question Attempts: Unlimited SAVE FOR LATER SUBMIT ANSWER SUBMIT ANSWER

Explanation / Answer

a)
given

n = 13.5 turns/cm = 1350 turns/m

R = 5.90 cm

I = 20.7 mA = 0.0207 A

let B_solenoid is the magnetic field produced by the solenoid and B_wire is the magnetic field produced by the wire and Bnet is the net magnetic field.

B_net*cos(theta) = B_solonoid ----(1)

B_net*sin(theta) = B_wire ----(2)

tan(theta) = B_wire/B_solonoid

B_solonoid*tan(theta) = B_wire

let r is the radial distance where we get the desired magnetic field.

magnetic field due to solenoid, B_solenoid = mue*n*I

= 4*pi*10^-7*1350*0.0207

= 3.51*10^-5 T

B_wire = mue*I'/(2*pi*r)

B_solonoid*tan(theta) = mue*I'/(2*pi*r)

r = mue*I'/(2*pi*B_solonoid*tan(theta))

= 4*pi*10^-7*8.52/(2*pi*3.51*10^-5*tan(56.5))

= 0.0321 m

= 3.21 cm <<<<<<<----------------Answer

b)

From equation 1

Bnet = B_solonoid/cos(theta)

= 3.51*10^-5/cos(56.5)

= 6.36*10^-5 T <<<<<<<----------------Answer

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