Question

The magnetic field at the center of a 1.5-cm-diameter loop is 2.7mT. Part A: What is the current in the loop? Part B: A long straight wire carries the same current you found in part A. At what distance from the wire is the magnetic field 2.7mT? Show work please

Answer 1

Concepts and reason

The main concept of this question is magnetic field due to a current carrying wire.

Initially, use the formula of the magnetic field due to current carrying circular loop to find the current flowing in it.

For the distance from the wire, use the formula of magnetic field due to straight current carrying wire.

Fundamentals

Consider a circular wire of radius r carrying current I through it.

The magnetic field B at the center of circular loop is given as follows:

$B = \frac{{{\mu ^\circ }}}{{4\pi }}\frac{{2\pi I}}{r}$

Here, ${\mu ^\circ }$ is the permeability of free space.

The expression for magnetic field ${B_1}$ due to long straight current carrying wire is,

${B_1} = \frac{{{\mu ^\circ }}}{{4\pi }}\frac{{2I}}{x}$

Here, ${\mu ^\circ }$ is the permeability of free space, I is the current flowing through it and x is the distance of field point from the wire.

(a)

The expression to calculate the current flowing through the circular loop is

$I = \frac{{4\pi }}{{{\mu ^\circ }}}\frac{{Br}}{{2\pi }}$

Substitute $4\pi \times {10^{ - 7}}{\rm{ T}} \cdot {\rm{m/A}}$ for ${\mu ^\circ }$ , $\frac{{1.5}}{2}{\rm{ cm}}$ for r and $2.7{\rm{ mT}}$ for B as follows :

$\begin{array}{c}\\I = \frac{{4\left( {3.14} \right)\left( {2.7{\rm{ mT}}} \right)\left( {\frac{{1 \times {{10}^{ - 3}}{\rm{ T}}}}{{1{\rm{ mT}}}}} \right)\left( {\frac{{1.5}}{2}{\rm{ cm}}} \right)\left( {\frac{{1 \times {{10}^{ - 2}}{\rm{ m}}}}{{1{\rm{ cm}}}}} \right)}}{{\left( {4\pi \times {{10}^{ - 7}}{\rm{ T}} \cdot {\rm{m/A}}} \right)2\left( {3.14} \right)}}\\\\ = 32.2{\rm{ A}}\\\end{array}$

(B)

The distance of the field point from the current carrying wire is

$x = \frac{{{\mu ^\circ }}}{{4\pi }}\frac{{2I}}{{{B_1}}}$

Substitute $4\pi \times {10^{ - 7}}{\rm{ T}} \cdot {\rm{m/A}}$ for ${\mu ^\circ }$ , ${\rm{32}}{\rm{.2 A}}$ for I and $2.7{\rm{ mT}}$ for B as follows :

$\begin{array}{c}\\x = \frac{{\left( {4\pi \times {{10}^{ - 7}}{\rm{ T}} \cdot {\rm{m/A}}} \right)}}{{4\left( {3.14} \right)}}\frac{{2\left( {32.2{\rm{ A}}} \right)}}{{\left( {2.7{\rm{ mT}}} \right)\left( {\frac{{1 \times {{10}^{ - 3}}{\rm{ T}}}}{{1{\rm{ mT}}}}} \right)}}\\\\ = 2.39 \times {10^{ - 3}}{\rm{ m}}\\\end{array}$

Ans: Part A

The magnitude of current flowing through the wire is $32.2{\rm{ A}}$ .

Part B

The distance of field point from the long straight current carrying wire is $2.39 \times {10^{ - 3}}{\rm{ m}}$ .