91影视

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

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

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

(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 \({\rm{49}}{\rm{.4}}\;{\rm{V/m}}\).