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White raven [17]
3 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?
Physics
1 answer:
sesenic [268]3 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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8 0
3 years ago
An electric field of 4.0 μV/m is induced at a point 2.0 cm from the axis of a long solenoid (radius = 3.0 cm, 800 turns/m). At w
vagabundo [1.1K]

Answer:

The rate of current in the solenoid  is 0.398 A/s

Explanation:

Given that,

Electric field E = 4.0\ \mu V/m

Distance = 2.0 cm

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We need to calculate the rate of current

Using formula of electric field for solenoid

E = \dfrac{x}{2}\mu_{0}n\dfrac{dI}{dt}

Where, x = distance

n = number of turns per unit length

E = electric field

r = radius

Put the value into the formula

4.0\times10^{-6}=\dfrac{2.0\times10^{-2}}{2}\times4\pi\times10^{-7}\times800\times\dfrac{dI}{dt}

\dfrac{dI}{dt}=\dfrac{4.0\times10^{-6}\times2}{2.0\times10^{-2}\times4\pi\times10^{-7}\times800}

\dfrac{dI}{dt}=0.397\ A/s

Hence, The rate of current in the solenoid  is 0.398 A/s.

6 0
4 years ago
Co nazywamy mitem? Dlaczego starożytni tworzyli mity?
Oxana [17]

Answer:

I dont understand polish sorry man

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3 years ago
Read 2 more answers
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