Question

(a) What is the direction of the force on a wire carrying a current due east in a location where the Earth’s field is due north? Both are parallel to the ground. (b) Calculate the force per meter if the wire carries \({\rm{20}}{\rm{.0 A}}\) and the field strength is\({\rm{3}}{\rm{.00 \times 10 - 5 T}}\). (c) What diameter copper wire would have its weight supported by this force? (d) Calculate the resistance per meter and the voltage per meter needed.

Short answer

a) The direction of movement is east and the magnetic field is north.

b) The force per meter = \({\rm{6}}{\rm{.00}} \times {\rm{1}}{{\rm{0}}^{{\rm{ - 4}}}}\;{\rm{N/m}}\).

c) Diameter is \({\rm{9}}{\rm{.42}} \times {\rm{1}}{{\rm{0}}^{{\rm{ - 3}}}}\;{\rm{m}}\).

d) Resistance is \({\rm{2}}{\rm{.47}}\;{\Omega \mathord{\left/ {\vphantom {\Omega {\rm{m}}}} \right. \\} {\rm{m}}}\) and voltage is \({\rm{49}}{\rm{.4}}\;{\rm{V/m}}\).

Step-by-step solution

Define force

The push or pull on a mass item that causes it to alter velocity is defined as a force.

Evaluating the direction of the force

(a)

If a person holds a current-carrying wire, the thumb depicts the current direction and the wrapped finger shows the magnetic line of force, according to Maxwell's right-hand rule.

The direction of movement is east, so the right thumbs point east, and the magnetic field is north, so the finger points up and the palm points outside of the page.

Evaluating the force per meter

(b)

In this current\({\rm{I}} = {\rm{20}}\;{\rm{A}}\), Magnetic field strength =\({\rm{3}}{\rm{.00}} \times {\rm{1}}{{\rm{0}}^{{\rm{ - 5}}}}\;{\rm{T}}\)and angle\({\rm{\theta }} = {\rm{9}}{{\rm{0}}^{\rm{o}}}\).

Magnetic force is,

\(\begin{aligned}{\rm{F}} &= {\rm{ILB}}\\\frac{{\rm{F}}}{{\rm{I}}} &= {\rm{IBsin\theta }}\\\frac{{\rm{F}}}{{\rm{I}}} &= {\rm{20}}\;{\rm{A}} \times \left( {{\rm{3}}{\rm{.00}} \times {\rm{1}}{{\rm{0}}^{{\rm{ - 5}}}}\;T} \right) \times {\rm{sin}}\left( {{\rm{9}}{{\rm{0}}^{\rm{o}}}} \right)\\\frac{{\rm{F}}}{{\rm{I}}} = {\rm{6}}{\rm{.00}} \times {\rm{1}}{{\rm{0}}^{{\rm{ - 4}}}}\;{\rm{N/m}}\end{aligned}\)

Hence, the force per meter = \({\rm{6}}{\rm{.00}} \times {\rm{1}}{{\rm{0}}^{{\rm{ - 4}}}}\;{\rm{N/m}}\).

Evaluating the diameter

(c)

Density of copper wire is,

\(\begin{aligned}{\rm{\rho }}& = \frac{{\rm{F}}}{{{\rm{\pi }}{{\rm{r}}^{\rm{2}}}{\rm{gl}}}}\\{\rm{r}} &= \sqrt {\frac{{\rm{F}}}{{{\rm{\pi \rho gl}}}}} \end{aligned}\)

Substituting the values in the above equation,

\(\begin{aligned}{\rm{r}} &= \sqrt {\frac{{{\rm{6}}{\rm{.00}} \times {\rm{1}}{{\rm{0}}^{{\rm{ - 4}}}}\;{\rm{N/m}}}}{{\left( {{\rm{8}}{\rm{.80}} \times {\rm{1}}{{\rm{0}}^{\rm{3}}}\;{{{\rm{kg}}} \mathord{\left/ {\vphantom {{{\rm{kg}}} {{{\rm{m}}^{\rm{3}}}}}} \right. \\} {{{\rm{m}}^{\rm{3}}}}}} \right) \times {\rm{\pi }} \times {\rm{9}}{\rm{.8}}\;{{\rm{m}} \mathord{\left/ {\vphantom {{\rm{m}} {{{\rm{s}}^{\rm{2}}}}}} \right. \\} {{{\rm{s}}^{\rm{2}}}}}}}} \\ &= {\rm{4}}{\rm{.71}} \times {\rm{1}}{{\rm{0}}^{{\rm{ - 5}}}}\;{\rm{m}}\end{aligned}\)

Then diameter is,

\(\begin{aligned}{\rm{d}} &= {\rm{2}} \times \left( {{\rm{4}}{\rm{.71}} \times {\rm{1}}{{\rm{0}}^{{\rm{ - 5}}}}\;{\rm{m}}} \right)\\ &= {\rm{9}}{\rm{.42}} \times {\rm{1}}{{\rm{0}}^{{\rm{ - 3}}}}\;{\rm{m}}\end{aligned}\)

Therefore, the diameter is \({\rm{9}}{\rm{.42}} \times {\rm{1}}{{\rm{0}}^{{\rm{ - 3}}}}\;{\rm{m}}\).

Evaluating the resistance and voltage

(d)

Resistance is calculated as,

\(\begin{aligned}\frac{{\rm{R}}}{L}{\rm{ = }}\frac{{\rm{\rho }}}{{\rm{A}}}\\\frac{{\rm{R}}}{L}{\rm{ = }}\frac{{\rm{\rho }}}{{{\rm{\pi }}{{\rm{r}}^{\rm{2}}}}}\end{aligned}\)

\(\begin{aligned}\frac{{\rm{R}}}{L} = \frac{{{\rm{1}}{\rm{.72}} \times {\rm{1}}{{\rm{0}}^{{\rm{ - 8}}}}\;\Omega \cdot {\rm{m}}}}{{{\rm{\pi }} \times {{{\rm{(4}}{\rm{.71}} \times {\rm{1}}{{\rm{0}}^{{\rm{ - 5}}}}\;{\rm{m)}}}^{\rm{2}}}}}\\ = {\rm{2}}{\rm{.47}}\;{\Omega \mathord{\left/ {\vphantom {\Omega {\rm{m}}}} \right. \\} {\rm{m}}}\end{aligned}\)

Voltage is calculated as,

\({\rm{V = IR}}\)

\(\begin{aligned}{\rm{V}} &= {\rm{2}}{\rm{.47}}\;{\Omega \mathord{\left/ {\vphantom {\Omega m}} \right. \\} m} \times L \times {\rm{20}}{\rm{.0}}\;{\rm{A}}\\\frac{{\rm{V}}}{L} &= {\rm{49}}{\rm{.4}}\;{\rm{V/m}}\end{aligned}\)

Therefore, resistance is \({\rm{2}}{\rm{.47}}\;{\Omega \mathord{\left/ {\vphantom {\Omega {\rm{m}}}} \right. \\} {\rm{m}}}\) and voltage is \({\rm{49}}{\rm{.4}}\;{\rm{V/m}}\).