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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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Explanation:

From the question we are told that

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The kinetic energy of the satellite is mathematically represented as

       KE  =  \frac{1}{2} * mv^2

where v is the speed of the satellite which is mathematically represented as

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=>  v^2  =  \frac{GM }{r}

substituting this into the equation

      KE  =  \frac{ 1}{2} *\frac{GMm}{r}

Now the gravitational force of the planet is mathematically represented as

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Where M is the mass of the planet and  m is the mass of the satellite

 Now looking at the formula for KE we see that we can represent it as

     KE  =  \frac{ 1}{2} *[\frac{GMm}{r^2}] * r

=>    KE  =  \frac{ 1}{2} *F_g * r

substituting values

       KE  =  \frac{ 1}{2} *6600 * 2.3*10^{7}

         KE  =  7.59  *10^{10} \  J

 

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