10. 1/3 points I Previous Answers $85 30 P30 My Notes Ask Your A superconducting
ID: 2305894 • Letter: 1
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10. 1/3 points I Previous Answers $85 30 P30 My Notes Ask Your A superconducting solenoid is meant to generate a magnetic field of 19.0 T. (a) If the solenoid winding has 2050 turns/m, what current is required? 7379x kA (b) What force per unit length is exerted on the windings by this magnetic field? 1402kN/m Drection 11. -1 points 585 31.P01 My Notes Ask Your A 55 turn rectangular coil of dimensions 5.00 cm x 10.0 cm is allowed to fall from a position where B 0 to a new position where B 0.470 T and is directed perpendicular to the plane of the coil. Calculate the magnitude of the average emf induced in the coil if the displacement occurs in 0.160 s mV 12 -/1 points 585 31 P20 Consider the arrangement shown in Figure P31.20. Assume that R 3.00 a, I = 1.20 m, and a uniform 2.50 T magnetic field is directed into the page. At what speed should the bar be moved to produce a current of 0.500 A in the resistor? My Notes Ask Your m/s Figure P31.20Explanation / Answer
10. given magnetic field, B = 19 T
a. number of windings per unit length, n/L = 2050 turns per m
current = i
now,
magnetic field inside superconducting solenoid is given by
B = mu*ni/L
hence
19 = 4*pi*10^-7*2050*i
i = 7375.472 A
b. force per unit length on the windings due to this magnetic field = F
F = B*i = 140.13398647 kN/m
11. numebr of turns, N = 55
dimesions of the rectangular coil, a x b = 5 x 10 cm
Bi = 0 T
Bf = 0.47 T
dt = 0.16 s
direction of magentic field perpendicular to the loop
hence
net induced flux
phi = ndB*ab
emf = n*dB*ab/dt
emf = 55*(0.47 - 0)*0.05*0.1/0.16
emf = 807.8125 mV
12. for the given arrangement
R = 3 ohms
l = 1.2 m
B = 2.5 T
speed of bar = v
current, i = 0.5 A
hence
EMF = iR = V
V = 1.5 V
but V = d(phi)/dt = B*v*l
hence
v = V/Bl = 1.5/2.5*1.2 = 0.55 m/s
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