Chapter 20, Problem 8P

To determine

The current carried by the wire.

To identify: Whether it is feasible or not.

Answer to Problem 8P

The current in the wire is 1400 A which is not feasible.

Explanation of Solution

Given:

The diameter of the copper wire is 1mm floating due to the Earth’s magnetic field $\overrightarrow{B}=5\times {10}^{-5} \text{T}$ . The density of the copper wire is $8.9\times {10}^3 {\text{km/m}}^{\text{3}}$

Formula used:

The magnetic force is calculated as

  $F_{mag}=mg=I(l\times B)$

Calculation:

The magnetic force applied in the wire must be equal in magnitude with the gravitational force applied to it. Since the wire is perpendicular to the Earth’s magnetic field thus angle of inclination would be ${90}^{\circ }$ . Now, using the density relation to calculate the mass of the copper wire

  $m=\rho \pi {\left(\frac{d}{2}\right)}^2$

Thus, the current in the copper wire is

  $\begin{array}{l}I=\frac{\rho \pi d^{2}g}{4B}\\ I=\frac{\text{(8}{\text{.9×10}}^{\text{3}}{\text{kg/m}}^{\text{3}}\text{)}\pi {{\text{(1×10}}^{\text{-3}}\text{m)}}^{\text{2}}\text{(9}{\text{.8m/s}}^{\text{2}}\text{)}}{{\text{4(5×10}}^{\text{-5}}\text{T)}}\\ I=1400 \text{A}\end{array}$

The current in the copper wire is way too high. Before reaching this magnitude of current, Joule heating in the thin copper wire would probably melt it. Thus, this answer is not feasible.

Conclusion:

Therefore, the current in wire is 1400 A and the answer is not feasible.