(12 points) A 15.0cm long horizontal (thick) wire of mass 15.Og is placed betwee
ID: 1285179 • Letter: #
Question
(12 points) A 15.0cm long horizontal (thick) wire of mass 15.Og is placed between two thin, vertical conductors, and a uniform magnetic field of 1 .0T acts perpendicular into the page. The wire is free to move vertically without friction on the two vertical conductors, and it would slide down if there were no current through it. What current should be sent through the wire, in which direction (left or right) to levitate the wire? 6. (10 points) A parallel-plate capacitor in air has a plate separation of 1.50cm and a plate area of 25.0 cm^2. The plates are charged to a potential difference of 250V and disconnected from the source. The space between the plates is then filled with distilled water, a dielectric with K=80. Find the a. capacitance before and after the water is added. b. the potential difference after the water is added, and the change in stored energy in the capacitor from before to after the water is added. . (20 points) The switch in the circuit below has been closed a long time, and a constant current flows. Use the voltage of the battery is 20V, R1 is 3.0k om. R2 is 8.0 k om , C1 is 3.7 nF and C2 is 8.2nF. a. What are the voltage drops across Cl and C2? b. How much charge (Ql and Q2) is stored on Cl and C2? The switch is opened at some t=0. After a long time, what are the new charges on the capacitors??Explanation / Answer
1)
Fg = FB
m*g = B*I*L*sin(90)
I = m*g/(B*L)
= 15*10^-3*9.8/(1*0.15)
= 0.98 A
direction of I should be towards right
2)
a) C = A*epsilon/d
= 25*10^-4*8.854*10^-12/1.5*10^-2
= 1.476*10^-12 F
C' = k*C
= 80*1.476
= 118*10^-12 F
b) here charge remainsa same.
Q' = Q
C'*V' = C*V
V' = V*(C/C')
= V/80
= 250/80
= 3.125 volts
c) delta U = Q^2/(2*C) - Q^2/(2*C')
= (Q^2/2*C)*(1 - 1/80)
= 0.5*C*V^2*(1 - 1/80)
= 0.5*1.476*10^-12*250^2*(1 - 1/80)
= 4.55*10^-8 J
3)
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