Chapter 20, Problem 31P

To determine

The maximum allowable current in the cable.

Answer to Problem 31P

The maximum allowable current in the wire is 2.5 A.

Explanation of Solution

Given:

Distance from the cable is 1.00 m. Accuracy of the experiment must be in the limits of $\pm 1.0\%$ .

Formula used:

The magnetic field due to the current in the wire is given by

  $B=\frac{{\mu }_{0}I}{2\pi r}$

Calculation:

As per the condition, the maximum allowable current in the wire up to the limits of accuracy must be

  $\begin{array}{l}{B}_{cable}=\frac{{\mu }_{0}I}{2\pi r}\le 0.01B_{Earth}\\ I\le \frac{2\pi r(0.01B_{Earth})}{{\mu }_0}=\frac{\text{2π(1m)(0}\text{.01)(0}{\text{.5×10}}^{\text{-4}}\text{T)}}{{\text{4π×10}}^{\text{-7}}\text{ T-m/A}}\\ I\le 2.5\text{A}\end{array}$

Conclusion:

Thus, the maximum allowable current in the wire is 2.5 A