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BlackZzzverrR [31]
3 years ago
14

A wire with length 3.5 m and mass 0.03 kg is aligned horizontally and connected by conducting loops to two vertical frictionless

metal rods. The rod is oriented in the East-West direction. What is the magnitude of the current that has to be run through the wire so that it hovers in place without sliding up or down? You may ignore magnetic forces caused by the vertical wires. Earth's magnetic field: 0.00005 Tesla or 0.5 gauss. Earth's gravity: 9.8 N/kg
Physics
1 answer:
Sindrei [870]3 years ago
6 0

Answer:

Current flowing in the wire is 1680 A

Explanation:

It is given length of wire l = 3.5 m

Mass of the wire m = 0.03 kg

Magnetic field B = 0.00005 Tesla

Acceleration due to gravity g=9.8m/sec^2

Mg force acting on the wire will be equal to Lorentz force acting on the wire.

Therefore mg=IBl

0.03\times 9.8=I\times 0.00005\times 3.5

I=1680A

Therefore current flowing in the wire is 1680 A

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y=y_{0} +v_{0} *t+\frac{1}{2} *g*t^{2} \\where:\\y= elevation [m]\\y_{0}=initial height [m]\\v_{0}= initial velocity [m/s] =41.67[m/s]\\t = time passed [s]\\g= gravity [m/s^2]=9.81[m/s^2]\\Now replacing:\\y=0+41.67 *(2)-\frac{1}{2} *(9.81)*(2)^{2} \\\\y=63.72[m]\\

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v_{f} =v_{o} +g*t\\replacing:\\v_{f} =41.67-(9.81)*(2)\\\\v_{f}=22.05[m/s]

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y=y_{o} +v_{o} *t-\frac{1}{2} *g*t^{2} \\y=63.72 +22.05 *t-\frac{1}{2} *(9.81)*t^{2} \\\\y=63.72 +22.05 *t-4.905*t^{2} \\

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Now we can establish with the conditions of the ball launched by David a new equation for y (elevation) in function of t, then we match these equations and find time t

y=y_{o} +v_{o} *t+\frac{1}{2} *g*t^{2} \\where:\\v_{o} =55.56[m/s] = initial velocity\\y_{o} =0[m]\\now replacing\\63.72 +22.05 *t-(4.905)*t^{2} =0 +55.56 *t-(4.905)*t^{2} \\63.72 +22.05 *t =0 +55.56 *t\\63.72 = 33.51*t\\t=1.9[s]

Then the time when both balls are going to be the same height will be when 2 [s] plus 1.9 [s] have passed after David throws the ball.

Time = 2 + 1.9 = 3.9[s]

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