Question

a. Derive an expression for the magnetic field strength at distance d from the center of a straight wire of finite length l that carries current I.

b. Determine the field strength at the center of a current carrying square loop having sides of length 2R.

c. Compare your answer to part b to the field at the center of a circular loop of diameter 2R. Do so by computing the ratio $\raisebox{1ex}{$B_{\text{square}}$}\!\left/ \!\raisebox{-1ex}{$B_{\text{circle}}$}\right.$.

Short answer

(a)$B=\begin{array}{rcl}& & \frac{{\mu }_0}{4\pi }\end{array}\begin{array}{rcl}& & \frac{I}{d}\end{array}\begin{array}{rcl}& & \left[\sin {\phi }_1+\sin {\phi }_2\right]\end{array}$.

(b)$B_{\text{square}}=\frac{2{\mu }_{0}I}{\pi R}$.

(c)$\left(\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$\pi $}\right.\right)$.

Step-by-step solution

Part (a): Step 1. Given information

The length of the wire is $l$, the current carrying by the wire is$I$and the distance of the observation point from the center of the wire is $d$.

Part (a): Step 2. Calculation

Let's consider the following diagram.

Let's assume that AB is a straight wire of length $L$and carrying current $I$. Also consider P be any point which is at a distance $d$from the wire where the magnetic field is to be calculated. Also consider an elemental length $d\stackrel{\rightharpoonup }{l}$on the wire which makes an angle $\theta$with respect to the position vector $\stackrel{\rightharpoonup }{r}$from the point P.

From the diagram, it can be written that

$\begin{array}{rcl}EG& =& EF\sin \theta \\ & =& dl\sin \theta ................................\left(1\right)\end{array}$

Also, from the diagram it can be written that,

$\begin{array}{rcl}EG& =& EPd\phi \\ & =& rd\phi ............................\left(2\right)\end{array}$

Equate equation (1) and (2) and simplify to obtain the expression for the elemental length.

$\begin{array}{rcl}dl\sin \theta & =& rd\phi \\ dl& =& \frac{rd\phi }{\sin \theta }.............................\left(3\right)\end{array}$

From $∆EQP$, it can be written that

$r=\frac{d}{\cos \phi }...........................\left(4\right)$

From Biot-Savart's law, the magnetic field at point P due to the elemental length is given by

role="math" localid="1650067905016" $dB=\frac{{\mu }_0}{4\pi }\frac{Idl\sin \theta }{r}...........................\left(5\right)$

Here, $dB$is the infinitesimal magnetic field and ${\mu }_0$is the permeability of free space.

Substitute the expression for $dl$and $r$from equation (3) and (4) respectively into equation (5) and simplify to obtain the infinitesimal magnetic field.

$\begin{array}{rcl}dB& =& \frac{{\mu }_0}{4\pi }\frac{I\left({\displaystyle \frac{rd\phi }{\sin \theta }}\right)\sin \theta }{r^2}\\ & =& \frac{{\mu }_0}{4\pi }\frac{Id\phi }{r}\\ & =& \frac{{\mu }_0}{4\pi }\frac{Id\phi }{\left({\displaystyle \frac{d}{\cos \phi }}\right)}\\ & =& \frac{{\mu }_0}{4\pi }\frac{I\cos \phi d\phi }{d}............................\left(6\right)\end{array}$

The formula to calculate the net magnetic field at P is given by

$B={\int }_{-{\phi }_1}^{{\phi }_2}dB.........................\left(7\right)$

Substitute the expression for $dB$from equation (6) into equation (7) and simplify to obtain the required magnetic field.

localid="1649562827741" $\begin{array}{rcl}B& =& {\int }_{-{\phi }_1}^{{\phi }_2}\frac{{\mu }_0}{4\pi }\frac{I\cos \phi d\phi }{d}\\ & =& \frac{{\mu }_0}{4\pi }\frac{I}{d}{\int }_{-{\phi }_1}^{{\phi }_2}\cos \phi d\phi \\ & =& \frac{{\mu }_0}{4\pi }\frac{I}{d}\left[\sin {\phi }_1+\sin {\phi }_2\right]\end{array}$

Part (a): Step 3. Final answer

The required magnetic field is given by$B=\begin{array}{rcl}& & \frac{{\mu }_0}{4\pi }\end{array}\begin{array}{rcl}& & \frac{I}{d}\end{array}\begin{array}{rcl}& & \left[\sin {\phi }_1+\sin {\phi }_2\right]\end{array}$.

Part (b): Step 1. Given information

The square loop has each side length of$2R$.

Part (b): Step 2. Calculation

A square loop can be thought of made up of four individual wires carrying the same current.

The formula to calculate the magnetic field $B$at a distance $r$due to a current carrying wire is given by

$B=\frac{{\mu }_{0}I}{2\pi r}.............\left(8\right)$

As the square loop has each side length of $2R$, it can be calculated from basic geometry that the distance of the center of the square from the center of any side is $R$.

The net magnetic field $\left(B_{\text{square}}\right)$at the center of the square is, then, given by

role="math" $B_{\text{square}}=4B.......................\left(8\right)$

Substitute the expression for $B$from equation (7) and $R$for $r$to obtain the required magnetic field.

$\begin{array}{rcl}{B}_{\text{square}}& =& 4\frac{{\mu }_{0}I}{2\pi R}\\ & =& \frac{2{\mu }_{0}I}{\pi R}\end{array}$

Part (b): Step 3. Final answer

The required magnetic field is given by$B_{\text{square}}=\frac{2{\mu }_{0}I}{\pi R}$.

Part (c): Step 1. Given information

The diameter of the circular loop is$2R$.

Part(c): Step2. Calculation

The formula to calculate the magnetic field $B_{\text{circle}}$at the center of a circular loop of diameter $2R$is given by

$B_{\text{circle}}=\frac{{\mu }_{0}I}{2R}.......................\left(9\right)$

Divide equation (8) by equation (9) and simplify to obtain the required ratio.

$\begin{array}{rcl}\frac{B_{\text{square}}}{B_{\text{circle}}}& =& \frac{\frac{2{\mu }_{0}I}{\pi R}}{\frac{{\mu }_{0}I}{2R}}\\ & =& \frac{4}{\pi }\end{array}$

Part (c): Step 3. Final answer

The required ratio is$\left(\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$\pi $}\right.\right)$.