Question

A closely wound, circular coil with radius 2.10cm has 830 turns.

A. What must the current in the coil be if the magnetic field at the center of the coil is 5.0010^-2

B. At what distance x from the center of the coil, on the axis of the coil, is the magnetic field half its value at the center?

Answer 1

Concepts and reason

The concepts required to solve this problem is magnetic field due to a circular coil.

Initially, use the expression of the magnetic field at the center of the circular coil and rearrange it to find the expression of current flowing in the coil. Then, substitute the values in the expression to find the value of current.

Then, use the expression of the magnetic field at the point on the axis of the circular coil and substitute the values to find the magnetic field at that point.

Fundamentals

The expression of the magnetic field at the center of the circular coil is given as follows:

$B = \frac{{{\mu _{\rm{o}}}nI}}{{2r}}$

Here, ${\mu _{\rm{o}}}$ is the permeability of free space, n is the number of turns, I is the current, and r is the radius of the coil.

The expression of the magnetic field at a distance x from the center of the coil along the axis of the coil is as follows:

$B = \frac{{{\mu _{\rm{o}}}nI}}{2}\frac{{{r^2}}}{{{{\left( {{x^2} + {r^2}} \right)}^{\frac{3}{2}}}}}$

Here, ${\mu _{\rm{o}}}$ is the permeability of free space, n is the number of turns, I is the current, x is the distance of the point at which the magnetic field is to be found from the center of the coil, and r is the radius of the coil.

(a)

The expression of the magnetic field at the center of the circular coil is given as follows:

$B = \frac{{{\mu _{\rm{o}}}nI}}{{2r}}$

Here, ${\mu _{\rm{o}}}$ is the permeability of free space, n is the number of turns, I is the current, and r is the radius of the coil.

Rearrange the above expression for I.

$I = \frac{{2Br}}{{{\mu _{\rm{o}}}n}}$

Substitute $5 \times {10^{ - 2}}{\rm{ T}}$ for B, $2.10{\rm{ cm}}$ for r, $4\pi \times {10^{ - 7}}{\rm{ H/m}}$for ${\mu _{\rm{o}}}$, and 830 for n in the above expression.

$\begin{array}{c}\\I = \frac{{2\left( {5 \times {{10}^{ - 2}}{\rm{ T}}} \right)\left( {2.10{\rm{ cm}}} \right)}}{{\left( {4\pi \times {{10}^{ - 7}}{\rm{ H/m}}} \right)\left( {830} \right)}}\left( {\frac{{{{10}^{ - 2}}\,{\rm{m}}}}{{1{\rm{ cm}}}}} \right)\\\\ = 2.01{\rm{ A}}\\\end{array}$

(b)

The expression of the magnetic field $B'$ at a distance x from the center of the coil along the axis of the coil is as follows:

$B' = \frac{{{\mu _{\rm{o}}}nI}}{2}\frac{{{r^2}}}{{{{\left( {{x^2} + {r^2}} \right)}^{\frac{3}{2}}}}}$

Rearrange the above expression for x.

${x^2} = {\left( {\frac{{{\mu _{\rm{o}}}nI{r^2}}}{{2B'}}} \right)^{\frac{2}{3}}} - {r^2}$

The magnetic field at the new point is half of the magnetic field at the center of the coil. The new value of magnetic field is as follows:

$B' = \frac{B}{2}$

Substitute $5 \times {10^{ - 2}}{\rm{ T}}$ for B in the above expression.

$\begin{array}{c}\\B' = \frac{{5 \times {{10}^{ - 2}}{\rm{ T}}}}{2}\\\\ = 2.5 \times {10^{ - 2}}{\rm{ T}}\\\end{array}$

Substitute $2.5 \times {10^{ - 2}}{\rm{ T}}$ for $B'$, $2.10{\rm{ cm}}$ for r, $4\pi \times {10^{ - 7}}{\rm{ H/m}}$for ${\mu _{\rm{o}}}$, 2.01 A for I, and 830 for n in the expression ${x^2} = {\left( {\frac{{{\mu _{\rm{o}}}nI{r^2}}}{{2B}}} \right)^{\frac{2}{3}}} - {r^2}$.

$\begin{array}{c}\\{x^2} = {\left( {\frac{{{\mu _{\rm{o}}}nI{r^2}}}{{2B}}} \right)^{\frac{2}{3}}} - {r^2}\\\\{x^2} = {\left( {\frac{{\left( {4\pi \times {{10}^{ - 7}}{\rm{ H/m}}} \right)\left( {830} \right)\left( {2.01{\rm{ A}}} \right){{\left( {2.10{\rm{ cm}}} \right)}^2}}}{{2\left( {2.5 \times {{10}^{ - 2}}{\rm{ T}}} \right)}}{{\left( {\frac{{{{10}^{ - 2}}{\rm{ m}}}}{{1{\rm{ cm}}}}} \right)}^2}} \right)^{\frac{2}{3}}} - {\left( {2.10{\rm{ cm}}} \right)^2}{\left( {\frac{{{{10}^{ - 2}}{\rm{ m}}}}{{1{\rm{ cm}}}}} \right)^2}\\\\x = \left( {\sqrt {6.79 \times {{10}^{ - 4}}{\rm{ }}{{\rm{m}}^2} - 4.41 \times {{10}^{ - 4}}{{\rm{m}}^2}} } \right)\left( {\frac{{1{\rm{ cm}}}}{{{{10}^{ - 2}}{\rm{ m}}}}} \right)\\\\x = 1.54{\rm{ cm}}\\\end{array}$

Ans: Part a

The current in the coil is 2.01 A.

Part b

At a point 1.54 cm from the coil, the magnetic field is half of the value of magnetic field at the center.