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AleksandrR [38]

2 years ago

14

If an electric wire is allowed to produce a magnetic field no larger than that of the Earth (0.55 x 10-4 T) at a distance of 25

cm, what is the maximum current the wire can carry?

2 answers:

antiseptic1488 [7]2 years ago

7 0

we are given in the problem the following dimensions or specifications
B = 0.000055 T r = 0.25 m constant mu0 = 4*pi*10-7

The formula that is applicable from physics is
B = mu0*I/(2*pi*r) I = 2*B*pi*r/mu0 I = 68.75 Amperes

Serga [27]2 years ago

4 0

Answer:

68.75 A

Explanation:

B = 0.55 x 10^-4 T, r = 25 cm = 0.25 m

Let the current be i.

The magnetic field due to a current carrying straight wire at a distance r is given by

B = μ0 / 4π x 2 i / r

0.55 x 10^-4 = 10^-7 x 2 x i / 0.25

i = 68.75 A

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7nadin3 [17]

The answer for the following problem is explained below.

Therefore the volume charge density of a substance (ρ) is 0.04 × $10^{-1}$ C.

Explanation:

Given:

radius (r) =2.1 cm = 2.1 × $10^{-2}$ m

height (h) =8.8 cm = 8.8 × $10^{-2}$ m

total charge (q) =6.1× $10^{-7}$ C

To solve:

volume charge density (ρ)

We know;

<u> ρ =q ÷ v</u>

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volume of cylinder (v) = 122.23 × $10^{-6}$

<u> ρ =q ÷ v</u>

ρ = 6.1× $10^{-7}$ ÷ 122.23 × $10^{-6}$

<u>ρ = 0.04 × </u> $10^{-1}$ <u> C</u>

Therefore the volume charge density of a substance (ρ) is 0.04 × $10^{-1}$ C.

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2 years ago

A proposed space elevator would consist of a cable stretching from the earth's surface to a satellite, orbiting far in space, th

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Where,

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1 year ago

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