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Aleks [24]
3 years ago
13

How large a current would a very long, straight wire have to carry so that the magnetic field 2.20 cm from the wire is equal to

1.00 g (comparable to the earth's northward-pointing magnetic field)?
Physics
1 answer:
choli [55]3 years ago
6 0
Biot-Savart Law - Magnetic field around a straight conductor carrying a current I.

B = μ_0 I/2πr

where:
B is strength of magnetic field in T.
μ_0 is permeability of free space = 1.25664 x 10^ -6 T-m/A.
I is current in A.
r is distance from conductor in m.

<span>So I = 2πrB /μ_0</span>
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Answer:

a) \frac{F}{w} =2.347\times 10^{-6}\ N

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Explanation:

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a.

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\frac{F}{w} =\frac{23\times 10^{-10}}{9.8\times10^{-4}}

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The condition for the bee to hang is its weight must get balanced by the electric force acing equally in the opposite direction.

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F=9.8\times10^{-4}\ N

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