WA assignment 10 chapter X O webassign.ne Responses/last?dep 15940028 Q7 t/web/S
ID: 1599687 • Letter: W
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WA assignment 10 chapter X O webassign.ne Responses/last?dep 15940028 Q7 t/web/Student/Assignment 1010 points Previous Answers SerPSEB 30.P.041.W. My Note Ask Yo A single turn square loop of wire, 2.00 cm on eech edge, cerries a clockwise current of 0.210 A. The loop is inside a solenoid, ith the plane of the loop perpendicular to the magnetic field of the solenoid, The solenoid has 30.0 turns/cm clockwise of 15.0 A d the forc h side of the loop. HN direction directed away from the center (b) Find the magnitude of the torque acting on the loop. Need Help? Read i Submit Answer Save Progress I Practice Another Version -10 points Ser SEB 30.F.043.Roln. My Notes It is desired to construct a solenoid that will have a resistance of 5.70 n (at 20.0°C) and produce a magnetic field of 4.so x 10 T its center when it carries a current of 3.20 A. The solenoid is to be constructed from copper wire having a dia (a) If the radius of the solenoid is to be 1 det ber of tu (b) If the radius of the sol is to be 1.00 cm, determine the required length of the solenoid. Need Help? Read a Tutor Submit Assignment Save Assignment Progress Home My Assignments Extension Requ Activate Windows Go to Settings to activate Windows. 1:36 PM Ask me anything 3/20/2017Explanation / Answer
a)
R = 5.70
we have R = *L/A
So L = R*A/
L = (5.70**(0.5 x 10^-3m)^2) /(1.72 x 10^-8)
L = 2.60 x 10^2 m
L = 260 m
Each turn of the solenoid is d = *2*0.01m = 6.28 x 10^-2m
so there are 260 / (6.28 x 10^-2) = 4140 turns
b)
The magnetic field B = µo*n*i
where n = turns per unit length
so n = B/(µo*i) = (4.8 x 10^-2) / (4 x 10^-7 x 3.2) = 11943 turns per unit length
We have the number of turn = 4140 and the necessary turns per unit length is 11940
so the length of the solenoid must be 4140/11940 = 0.110m = 34.67 cm
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