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Expert-verifiedA surveyor \({\rm{100 m}}\) from a long straight \({\rm{200}}\;{\rm{kV}}\) DC power line suspects that its magnetic field may equal that of the Earth and affect compass readings. (a) Calculate the current in the wire needed to create a \({\rm{5}}{\rm{.00 \times 1}}{{\rm{0}}^{{\rm{ - 5}}}}{\rm{ T}}\) field at this distance. (b) What is unreasonable about this result? (c) Which assumption or premise is responsible?
A vector field that can interact with the moving charges and other magnetic substances is known as a magnetic field. Force produced between two objects due to magnetic interactions can be both attractive and repulsive.
(a)
The current is expressed in the terms of magnetic field by the relation: \(I = \frac{{2\pi rB}}{{{\mu _0}}}\).
Here, the value of \({{\rm{\mu }}_{\rm{0}}}\) is said to be the permeability of the free space.
\(\begin{aligned}{}I &= \frac{{2\pi rB}}{{{\mu _0}}}\\ &= \frac{{2\pi \times 100 {\rm{m}} \times \left( {5.00 \times {{10}^{ - 5}}} \right) {\rm{T}}}}{{\left( {4\pi \times {{10}^{ - 7}}} \right) {\rm{T}} \cdot {\rm{m/A}}}}\\ &= 2.5 \times {10^4}\;{\rm{A}}\\ &= 25\;{\rm{kA}}\end{aligned}\)
Therefore, current in the wire is: \({\rm{25 kA}}\).
(b)
The current evaluated in the part (a) is very large. The power is too high for standard transmission lines.
(c) The \({\rm{100 m}}\) distance is said to be unreasonably long for the measurement of magnetic field. Also parallel cables are used as transmission line, so that there is no net magnetic field produced, for DC power lines.
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