Work on a separate sheet. In order to get full credit show all your work. Credit

Question

Work on a separate sheet. In order to get full credit show all your work. Credit will be given to each step of the work A wire (mass = 50 g, length = 40 cm) is suspended horizontally by two vertical wires which conduct a current I = 8.0 A, as shown in the figure. The magnetic field in the region is into the paper and has a magnitude of 60 mT. What is the tension in either wire? The figure below shows a schematic representation of the apparatus that can be used to measure magnetic fields. A rectangular coil of wire contains N turns and has a width w. The coil is attached to one arm of a balance and is suspended between the poles of a magnet. The magnetic field is uniform and perpendicular to the plane of the coil. The system is first balanced when the current in the coil is zero. When the switch is closed and the coil carries a current I, a mass m must be added to the right side to balance the system. Find an expression for the magnitude of the magnetic field. (Use any variable or symbol stated above along with the following as necessary: g.) The pair of capacitors in the figure below are fully charged by a 12.0-V battery. The battery is disconnected, and the switch is then closed. After 1.0 ms has elapsed, how much charge remains on the 3.00-mu F capacitor? After 1.0 ms has elapsed, how much charge remains on the 2.00-mu F capacitor? What is the current in the resistor at this time?

Explanation / Answer

1) Tension is a force F= BIL
where F is force or tension
B= magnetic field density measured in Teslas (T) in this case 0.060 T
I= The current measured in amps A (in this case 8.0 A)
and obviously L is length of the wire
therefore, F= 0.060T* 8A*0.40m = 0.192 N and that's the tension

3) Time Constant = RC = 5*500*10^-6 = 2.5 ms

Net capacitance = C = 5.0 uF

Total charge = Qo = CV = 12*5 = 60 uC

Discharging, Q = Qo*exp(-t/RC) = 60*10^-6*exp(-1/2.5) = 40.2 uC

Now Voltage across 3.0 uF = Voltage across 2.0 uF

V3 = V2

Q3/C3 = Q2/C2 ========> Q3/3 = Q2/2 =========> Q3 = 1.5*Q2

Also Q = Q2 + Q3 = 2.5*Q2

40.2*10^-6 = 2.5*Q2 ======> Q2 = 16.1 uC ................. .(Answer of part b)

Q3 = 1.5*16.1 = 24.15 uC ...................(Answer of part a)

   c) Now V = Q2/C2 = 16.1/2 = 8.05 V

I = V/R = 8.05/500 = 16.1 mA

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