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The following data are the magnitude of the magnetic field at various distances from a long, straight, thick cylindrical copper cable that is carrying a large constant current, we expect that the quantity Bx from these data

will be constant, this quantity for each pair of the given data are 0.812, 1.00, 1.09, 1.13, 1.16 (all in the unit mT cm, we can see that this quantity is not constant as it should be, but also we can see that for the last three values the values are close to each other. We know that the magnetic field due to an along, straight, thick cylindrical copper cable is inversely proportional to r which is the distance from the center of the cable, but t is not the distance from the center of the cable, it is the distance from the outer surface of the cable, so r = t+R, where R is the radius of the cable. The magnetic field is, therefore,

$B=\frac{{渭}_{o}I}{2蟺(R+x)}$

Solving equation (1) for x we get,

$x=\left(\frac{{渭}_{o}I}{2蟺}\right)\frac{1}{B}鈭�R$

according to this equation, a graph between a and 1/B should be a straight line, with the slope of$\frac{u_{o}I}{2\mathrm{蟺}}$and y-intercept of - R. To make this plot,

Finally, we need to use the graph in part (b) to calculate I and the radius of the cable, the fitting equation is,

$x=\underset{\text{slope}}{(0.0013\mathrm{T}}鈰�\mathrm{cm})\frac{1}{B}鈭�1.1914\mathrm{cm}$

The slope is equal to localid="1668269168258" $\frac{u_{o}I}{2\mathrm{蟺}}$, then the current is,

$\begin{array}{c}I=\frac{2蟺\left(\text{slope}\right)}{{渭}_0}\\ =\frac{2蟺\left(1.3脳{10}^{鈭�5}\mathrm{T}鈰�\mathrm{m}\right)}{4蟺脳{10}^{鈭�7}\mathrm{T}鈰�\mathrm{m}/\mathrm{A}}\\ =65\mathrm{A}\\ I=65\mathrm{A}\end{array}$

the radius of the cable equals the y-intercept, that is,

R = 1.1914 cm