Question

You use a tesla meter (a Hall-effect device) to measure the magnitude of the magnetic field at various distances from a long, straight, thick cylindrical copper cable that is carrying a large constant current. To exclude the earth’s magnetic field from the measurement, you first set the meter to zero. You then measure the magnetic field Bat distances xfrom the surface of

the cable and obtain these data: x1cm2 2.0 4.0 6.0 8.0 10.0 B1mT2 0.406 0.250 0.181 0.141 0.116

(a) You think you remember from your physics course that the magnetic field of a wire is inversely proportional to the distance from the wire. Therefore, you expect that the quantity Bxfrom your data will be constant. Calculate Bxfor each data point in the table. Is Bxconstant for this set of measurements? Explain.

(b) Graph the data as xversus 1>B. Explain why such a plot lies close to a straight line.

(c) Use the graph in part (b) to calculate the current Iin the cable and the radius Rof the cable.

Short answer

a) $B=\frac{{\mu }_{o}I}{2\pi (R+x)}$

b) See the graph in step 2

c) l = 65A and R = 1.914 cm

Step-by-step solution

Solving part (a) of the problem.

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,

$B=\frac{{\mu }_{o}I}{2\pi (R+x)}$

Solving part (b) of the problem.

Solving equation (1) for x we get,

$x=\left(\frac{{\mu }_{o}I}{2\pi }\right)\frac{1}{B}-R$

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

Solving part (c) of the problem.

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

$x=\underset{\text{slope}}{(0.0013\mathrm{T}}\cdot \mathrm{cm})\frac{1}{B}-1.1914\mathrm{cm}$

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

$\begin{array}{c}I=\frac{2\pi \left(\text{slope}\right)}{{\mu }_0}\\ =\frac{2\pi \left(1.3\times {10}^{-5}\mathrm{T}\cdot \mathrm{m}\right)}{4\pi \times {10}^{-7}\mathrm{T}\cdot \mathrm{m}/\mathrm{A}}\\ =65\mathrm{A}\\ I=65\mathrm{A}\end{array}$

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

R = 1.1914 cm