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White raven [17]

2 years ago

11

Your friend wants to be magician and intends to use Earth’s magnetic field to suspend a current-carrying wire above the stage. H

e asks you to estimate the minimum current needed to suspend the wire just above Earth’s surface at the equator (where Earth’s magnetic field is horizontal). Assume the wire has a mass of 10 g, and length of 1.0 m. Would you advise him to proceed with his plans for this act?

1 answer:

sesenic [268]2 years ago

7 0

Answer:

$I=1960A$

Explanation:

Under this condition, . In order to suspend the wire, this magnetic force would have to be equal in magnitude to the gravitational force exerted by Earth on the wire the maximum force at angle

F=ILxB

Now to the suspend the wire so use the maximum force and solve to the current knowing the magnetic field of the earth

∑Fy=0

$F_{m}-F_{g}=0$

$I*L*\beta-m*g=0$

Solve to I current

$I=\frac{m*g}{L*\beta}$

$I=\frac{10x10^{-3}kg*9.8m/s^2}{1m*0.5x10{-4}T}$

$I=1960A$

I suggest do an toher act is really risk that current for an act

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Determine the index of refraction of glass that is struck by unpolarized light at 53.8 degrees and resulting in a fully polarize

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A uniform magnetic field is perpendicular to the plane of a circular loop of diameter 13 cm formed from wire of diameter 2.6 mm

I am Lyosha [343]

Answer:

Rate of change of magnetic field is $3.466\times 10^3T/sec$

Explanation:

We have given diameter of the circular loop is 13 cm = 0.13 m

So radius of the circular loop $r=\frac{0.13}{2}=0.065m$

Length of the circular loop $L=2\pi r=2\times 3.14\times 0.065=0.4082m$

Wire is made up of diameter of 2.6 mm

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Resistivity of wire $\rho =2.18\times 10^{-8}m$

Resistance of wire $R=\frac{\rho L}{A}=\frac{2.18\times 10^{-8}\times 0.4082}{5.30\times 10^{-6}}=1.67\times 10^{-3}ohm$

Current is given i = 11 A

So emf $e=11\times 1.67\times 10^{-3}=0.0183volt$

Emf induced in the coil is $e=-\frac{d\Phi }{dt}=-A\frac{dB}{dt}$

$0.0183=5.30\times 10^{-6}\times \frac{dB}{dt}$

$\frac{dB}{dt}=3.466\times 10^3=T/sec$

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