Plese solve all 3 qustion in these pictures. o Ask your Teacher points SerPsER 2
ID: 1535114 • Letter: P
Question
Plese solve all 3 qustion in these pictures. o Ask your Teacher points SerPsER 20 P00 The upper portion of the circuit in the figure below is fixed. The horizontal wire at the bottom has a mass of 10.0 g and is 5.00 cm long. This wire hangs in the gravitational field of the Earth from identical light springs connected to the upper portion of the circuit. The springs stretch o 600 cm under the weight of the wire, and the circuit has a total resistance of 12.0 n. when a magnetic field is turned on, directed out of the page, the springs stretch an additional o 360 cm. only the horizontal wire at the bottom of the crcuit is in the magnetic field. What is the of the magnetic field? 24.0 V 5.00 cm Need Help? LReson ISuomitAnsweil Save Progress Practice Anowher version My Notes o Ask Your reacher T o -a points seeps
Explanation / Answer
spring constant k = Fg/(2*y1) = mg/(2*y1)
K = (10*10^-3*9.8)/(2*0.6*10^-2) = 8.1 N/m
crrent in the lower wire I = E/R = 24/12 = 2 A
magnetic force Fb = I*L*B
elastic force Fe = 2k*y2
Fe = Fb
2*8.1*0.36*10^-2 = 2*0.05*B
B = 0.5832 T
(a)
kinetic energy K = (1/2)*m*v^2 = 5.08 MeV
(1/2)*1.67*10^-27*v^2 = 5.08*10^6*1.6*10^-19
vx = 3.2*10^7 m/s i
along horizontal
x = vx*t
t = x/vx = 1/(3.2*10^7) = 3.12*10^-8 s
magnetic force Fb = q*(v x B )
Fb = 1.6*10^-19*(3.2*10^7 i x 0.0496 k )
Fb = -2.54*10^-13 j N
acceleration ay = Fb/m = -2.54*10^-13/(1.6*10^-27) = -1.6*10^14 m/s^2 j
along vertical
vy = voy + ay*t
vy = 0 - 1.6*10^14*3.12*10^-8 = -5*10^6
alpha = tan^-1(vy/vx)
alpha = 8.87 degrees
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(b)
Py = m*vy = 1.67*10^-27*5*10^6 = 8.35*10^-21 kg m/s
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9)
magnetic force Fb = q*v*B
acceleration a = F/m = q*v*B/m
along perepndicular
y = voy*t + (1/2)*ay*t^2
y = 0 + (1/2)*q*v*B*t^2/m
0.15 = (1/2)*q*1.48*10^4*0.161*10^-3*1^2/(1.38*10^-3)
q = 174 uC