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Agata [3.3K]
3 years ago
7

A wire of radius 0.8 cm carries a current of 106 A that is uniformly distributed over its cross-sectional area. Find the magneti

c field B at a distance of 0.07 cm from the center of the wire.
Physics
1 answer:
Maurinko [17]3 years ago
5 0

Answer:

The magnetic field is B =  2.319 *10^{-3} \  T

Explanation:

From the question we are told that

   The radius of the wire is r = 0.8 \ cm = 0.008 \  m

    The current is I  =  106 \  A

    The position considered is  d = 0.07 cm = 0.0007 m

 Generally the magnetic field is mathematically represented as

         B =   \frac{\mu_o * I}{2\pi * \frac{r^2}{d} }

Here \mu_o is the permeability of free space with value 4\pi * 10^{-7} N/A^2

So    

        B =   \frac{ 4\pi * 10^{-7} * 106 }{2 * 3.142  * \frac{0.008^2}{0.0007} }

=>     B =  2.319 *10^{-3} \  T

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<span>cos(90 - theta) is just sin(theta) </span>
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<span>There are a few ways to attack this. Basically, you have to find the angle from the origin to the apogee (highest point) in the arc. Once we have that, we'll know what angle the momentum vector makes with the moment-arm because, at the apogee, we know that all of the motion is *horizontal*. </span>

<span>Okay, so let's get back to what we know: </span>

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<span>m is still m. </span>
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<span>That leaves us with phi, the angle the horizontal velocity vector makes with the moment arm. To find *that*, we need to know what the angle from the origin to the apogee is. We can find *that* by taking the arc-tangent of the slope, if we know that. Well, we know the "run" part of the slope (it's our "d" term), but not the rise. </span>

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<span>L = [ vi^2 * sin(2*theta) / 2g ] * m * [ vi * cos(theta) ] * [ cos( arctan( tan(theta) / 4 ) ) ] </span>
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