Question

a) What current is needed to generate the magnetic field strength of 5.0*10^-5 T at a point 2.3 cm from a long, straight wire?

b) What current is needed to generate the magnetic field strength of 5.0*10^-3 T at a point 2.3 cm from a long, straight wire?

c) What current is needed to generate the magnetic field strength of 1 T at a point 2.3 cm from a long, straight wire?

d) What current is needed to generate the magnetic field strength of 10 T at a point 2.3 cm from a long, straight wire?

Answer 1

Concepts and reason

Concepts here will revolve around magnetic field at a point from a straight line, varying magnetic fields, and current associated with it.

Relation between current and magnetic is obtained from Biot-Savart law, which can be varied from case to case to get results.

Fundamentals

The magnetic field at a point at a distance from current carrying conductor is given by

$B = \frac{{{\mu _0}I}}{{2\pi d}}$

Here $B$is the magnetic field, ${\mu _0}\left( { = 4\pi \times {{10}^{ - 7}}\;\frac{{\rm{H}}}{{\rm{m}}}} \right)$ is the permeability of free space, $I$is the associated current, $\pi \left( { = 3.14} \right)$is a constant and $d$is the distance of the point from the rod.

(a)

Rearrange the expression $B = \frac{{{\mu _0}I}}{{2\pi d}}$ for current (I),

$I = \frac{{2\pi Bd}}{{{\mu _0}}}$

The value of current due to the long straight wire can be calculated as,

$I = \frac{{2\pi Bd}}{{{\mu _0}}}$

Convert the unit of distance from cm to ${\rm{m}}$.

$\begin{array}{c}\\d = {\rm{2}}{\rm{.3}}\;{\rm{cm}}\\\\{\rm{ = 2}}{\rm{.3}}\,{\rm{cm}}\left( {\frac{{1{\rm{ m}}}}{{{\rm{100 cm}}}}} \right)\\\\ = 0.023\;{\rm{m}}\\\end{array}$

Substitute ${\rm{0}}{\rm{.023}}\;{\rm{m}}$for $d$,${\rm{5 \times 1}}{{\rm{0}}^{{\rm{ - 5}}}} {\rm{T}}$ for $B$,${\rm{4\pi \times 1}}{{\rm{0}}^{{\rm{ - 7}}}} {\rm{T}} \cdot {\rm{m}} \cdot {{\rm{A}}^{ - 1}}$ for ${\mu _0}$ in the expression of magnetic field due to straight wire.

$\begin{array}{c}\\I = \frac{{{\rm{2\pi }}\left( {{\rm{5 \times 1}}{{\rm{0}}^{{\rm{ - 5}}}} {\rm{T}}} \right)\left( {{\rm{0}}{\rm{.023}} {\rm{m}}} \right)}}{{\left( {{\rm{4\pi \times 1}}{{\rm{0}}^{{\rm{ - 7}}}} {\rm{T}} \cdot {\rm{m}} \cdot {{\rm{A}}^{ - 1}}} \right)}}\\\\ = 5.75 {\rm{A}}\\\end{array}$

(b)

The value of current due to the long straight wire can be calculated as,

$I = \frac{{2\pi Bd}}{{{\mu _0}}}$

Convert the unit of distance ${\rm{cm}}$to ${\rm{m}}$.

$\begin{array}{c}\\d = {\rm{2}}{\rm{.3}}\;{\rm{cm}}\\\\{\rm{ = 2}}{\rm{.3}}\,{\rm{cm}}\left( {\frac{{1{\rm{ m}}}}{{{\rm{100 cm}}}}} \right)\\\\ = 0.023\;{\rm{m}}\\\end{array}$

Substitute ${\rm{0}}{\rm{.023}}\;{\rm{m}}$for $d$, ${\rm{5 \times 1}}{{\rm{0}}^{{\rm{ - 3}}}} {\rm{T}}$for $B$,${\rm{4\pi \times 1}}{{\rm{0}}^{{\rm{ - 7}}}} {\rm{T}} \cdot {\rm{m}} \cdot {{\rm{A}}^{ - 1}}$ for ${\mu _0}$ in the expression of magnetic field due to straight wire.

$\begin{array}{c}\\I = \frac{{{\rm{2\pi }}\left( {{\rm{5 \times 1}}{{\rm{0}}^{{\rm{ - 3}}}} {\rm{T}}} \right)\left( {{\rm{0}}{\rm{.023}} {\rm{m}}} \right)}}{{\left( {{\rm{4\pi \times 1}}{{\rm{0}}^{{\rm{ - 7}}}} {\rm{T}} \cdot {\rm{m}} \cdot {{\rm{A}}^{ - 1}}} \right)}}\\\\ = 575 {\rm{A}}\\\end{array}$

(c)

The value of current due to the long straight wire can be calculated as,

$I = \frac{{2\pi Bd}}{{{\mu _0}}}$

Convert the unit of distance from cm to ${\rm{m}}$.

$\begin{array}{c}\\d = {\rm{2}}{\rm{.3}}\;{\rm{cm}}\\\\{\rm{ = 2}}{\rm{.3}}\,{\rm{cm}}\left( {\frac{{1{\rm{ m}}}}{{{\rm{100 cm}}}}} \right)\\\\ = 0.023\;{\rm{m}}\\\end{array}$

Substitute ${\rm{0}}{\rm{.023}}\;{\rm{m}}$for $d$, ${\rm{1}} {\rm{T}}$for$B$,${\rm{4\pi \times 1}}{{\rm{0}}^{{\rm{ - 7}}}} {\rm{T}} \cdot {\rm{m}} \cdot {{\rm{A}}^{ - 1}}$ for${\mu _0}$ in the expression of magnetic field due to straight wire.

$\begin{array}{c}\\I = \frac{{{\rm{2\pi }}\left( {{\rm{1}} {\rm{T}}} \right)\left( {{\rm{0}}{\rm{.023}} {\rm{m}}} \right)}}{{\left( {{\rm{4\pi \times 1}}{{\rm{0}}^{{\rm{ - 7}}}} {\rm{T}} \cdot {\rm{m}} \cdot {{\rm{A}}^{ - 1}}} \right)}}\\\\ = 115 \times {10^3} {\rm{A}}\\\end{array}$

(d)

The value of current due to the long straight wire can be calculated as,

$I = \frac{{2\pi Bd}}{{{\mu _0}}}$

Convert the unit of distance from cm to ${\rm{m}}$.

$\begin{array}{c}\\d = {\rm{2}}{\rm{.3}}\;{\rm{cm}}\\\\{\rm{ = 2}}{\rm{.3}}\,{\rm{cm}}\left( {\frac{{1{\rm{ m}}}}{{{\rm{100 cm}}}}} \right)\\\\ = 0.023\;{\rm{m}}\\\end{array}$

Substitute ${\rm{0}}{\rm{.023}}\;{\rm{m}}$for $d$,${\rm{10}} {\rm{T}}$for$B$,${\rm{4\pi \times 1}}{{\rm{0}}^{{\rm{ - 7}}}} {\rm{T}} \cdot {\rm{m}} \cdot {{\rm{A}}^{ - 1}}$ for${\mu _0}$ in the expression of magnetic field due to straight wire.

$\begin{array}{c}\\I = \frac{{{\rm{2\pi }}\left( {{\rm{10}} {\rm{T}}} \right)\left( {{\rm{0}}{\rm{.023}} {\rm{m}}} \right)}}{{\left( {{\rm{4\pi \times 1}}{{\rm{0}}^{{\rm{ - 7}}}} {\rm{T}} \cdot {\rm{m}} \cdot {{\rm{A}}^{ - 1}}} \right)}}\\\\ = 115 \times {10^4} {\rm{A}}\\\end{array}$

Ans: Part a

The value of current due to the long straight wire is $5.75{\rm{ A}}$.

Part b

The value of current due to the long straight wire is $575{\rm{ A}}$.

Part c

The value of current due to the long straight wire is $115 \times {10^3}\;{\rm{A}}$.

Part d

The value of current due to the long straight wire is $115 \times {10^{4\;}}\;{\rm{A}}$.