Question 14 Marks
Derive the formula for the magnetic field produced at a point on the axis of a current carrying circular loop.
Answer
View full question & answer→$\rightarrow$ As shown in the figure, a steady current I is flowing through a conducting loop of radius $R$.
$\rightarrow$ The loop is placed in such a way that it lies in the $y-z$ plane and the $X -$axis passing through its axis.
$\rightarrow$ A point $P$ lies at a distance $x$ on the $X -$axis from its origin. We want to calculate the magnetic field at the point $P .$
$\rightarrow$ Consider a current element $I d \vec{l}$ from the loop shown in figure. The magnitude of the magnetic field due to this element is,
$d B=\frac{\mu_0}{4 \pi} \cdot \frac{| I d \vec{l} \times \vec{r}|}{r^3}$
$\rightarrow$ But $I d \vec{l} \perp \vec{r}$ because $I d \vec{l}$ is in the $y z$ plane and the position vector $(\vec{r})$ is in $x y$ plane.
$\therefore d B=\frac{\mu_0}{4 \pi} \cdot \frac{ I d l r \sin 90}{r^3}$
$\therefore d B=\frac{\mu_0}{4 \pi} \cdot \frac{ I d l}{r^2}$
$\rightarrow$ From the figure, $r^2= R ^2+x^2$. Hence,
$d B=\frac{\mu_0}{4 \pi} \cdot \frac{ I d l}{\left( R ^2+x^2\right)}$
$\rightarrow$ The magnetic field has two components at point $P$
$(i)$ Perpendicular component $\left(d B_{\perp}=d B \sin \theta\right)$
$\rightarrow$ When the perpendicular components are summed to get the net magnetic field, they cancel each other and the result is zero
$(ii)$ Parallel component $\left(d B_{\|}=d B \cos \theta\right)$
$\rightarrow$ The parallel components are summed up to get the net magnetic field, so it can be obtained by integrating $d B_x=d B \cos \theta$ over the loop.
$\rightarrow \Rightarrow d B(x)=d B \cos \theta$
$\therefore d B(x)=\frac{\mu_0}{4 \pi} \cdot \frac{ I d l}{ R ^2+x^2} \cdot \cos \theta$
$(\because$ from equation $(3))$
$\rightarrow$ From the figure,
$\cos \theta =\frac{ R }{\left(x^2+ R ^2\right)^{\frac{1}{2}}}$
$\therefore d B(x) =\frac{\mu_0}{4 \pi} \cdot \frac{ I d l}{ R ^2+x^2} \cdot \frac{ R }{\left( R ^2+x^2\right)^{\frac{1}{2}}}$
$\therefore d B(x) =\frac{\mu_0}{4 \pi} \cdot \frac{ I d l \cdot R }{\left(R^2+x^2\right)^{\frac{3}{2}}}$
$\rightarrow$ The resultant magnetic field.
$B =\oint d B(x)=\frac{\mu_0 IR }{4 \pi\left( R ^2+x^2\right)^{\frac{3}{2}}} \oint d l$
$\therefore B =\frac{\mu_0 IR }{4 \pi\left( R ^2+x^2\right)^{\frac{3}{2}}}(2 \pi R )(\because \oint d l=2 \pi R )$
$\therefore B =\frac{\mu_0 IR ^2}{2\left( R ^2+x^2\right)^{\frac{3}{2}}}$
$\rightarrow$ In vector form,
$\overrightarrow{ B }=\frac{\mu_0 IR ^2}{2\left( R ^2+x^2\right)^{\frac{3}{2}}} \cdot \hat{i}$
$\rightarrow$ To obtain the magnetic field at the centre of the loop $x=0$
$\therefore B=\frac{\mu_0 IR ^2}{2 R ^3}=\frac{\mu_0 I }{2 R }$
$\rightarrow$ If there are $N$ turns, then
$\overrightarrow{ B }=\frac{\mu_0 NIR ^2}{2\left( R ^2+x^2\right)^{\frac{3}{2}}} \cdot \hat{i}$
$\rightarrow$ The loop is placed in such a way that it lies in the $y-z$ plane and the $X -$axis passing through its axis.
$\rightarrow$ A point $P$ lies at a distance $x$ on the $X -$axis from its origin. We want to calculate the magnetic field at the point $P .$
$\rightarrow$ Consider a current element $I d \vec{l}$ from the loop shown in figure. The magnitude of the magnetic field due to this element is,
$d B=\frac{\mu_0}{4 \pi} \cdot \frac{| I d \vec{l} \times \vec{r}|}{r^3}$
$\rightarrow$ But $I d \vec{l} \perp \vec{r}$ because $I d \vec{l}$ is in the $y z$ plane and the position vector $(\vec{r})$ is in $x y$ plane.
$\therefore d B=\frac{\mu_0}{4 \pi} \cdot \frac{ I d l r \sin 90}{r^3}$
$\therefore d B=\frac{\mu_0}{4 \pi} \cdot \frac{ I d l}{r^2}$
$\rightarrow$ From the figure, $r^2= R ^2+x^2$. Hence,
$d B=\frac{\mu_0}{4 \pi} \cdot \frac{ I d l}{\left( R ^2+x^2\right)}$
$\rightarrow$ The magnetic field has two components at point $P$
$(i)$ Perpendicular component $\left(d B_{\perp}=d B \sin \theta\right)$
$\rightarrow$ When the perpendicular components are summed to get the net magnetic field, they cancel each other and the result is zero
$(ii)$ Parallel component $\left(d B_{\|}=d B \cos \theta\right)$
$\rightarrow$ The parallel components are summed up to get the net magnetic field, so it can be obtained by integrating $d B_x=d B \cos \theta$ over the loop.
$\rightarrow \Rightarrow d B(x)=d B \cos \theta$
$\therefore d B(x)=\frac{\mu_0}{4 \pi} \cdot \frac{ I d l}{ R ^2+x^2} \cdot \cos \theta$
$(\because$ from equation $(3))$
$\rightarrow$ From the figure,
$\cos \theta =\frac{ R }{\left(x^2+ R ^2\right)^{\frac{1}{2}}}$
$\therefore d B(x) =\frac{\mu_0}{4 \pi} \cdot \frac{ I d l}{ R ^2+x^2} \cdot \frac{ R }{\left( R ^2+x^2\right)^{\frac{1}{2}}}$
$\therefore d B(x) =\frac{\mu_0}{4 \pi} \cdot \frac{ I d l \cdot R }{\left(R^2+x^2\right)^{\frac{3}{2}}}$
$\rightarrow$ The resultant magnetic field.
$B =\oint d B(x)=\frac{\mu_0 IR }{4 \pi\left( R ^2+x^2\right)^{\frac{3}{2}}} \oint d l$
$\therefore B =\frac{\mu_0 IR }{4 \pi\left( R ^2+x^2\right)^{\frac{3}{2}}}(2 \pi R )(\because \oint d l=2 \pi R )$
$\therefore B =\frac{\mu_0 IR ^2}{2\left( R ^2+x^2\right)^{\frac{3}{2}}}$
$\rightarrow$ In vector form,
$\overrightarrow{ B }=\frac{\mu_0 IR ^2}{2\left( R ^2+x^2\right)^{\frac{3}{2}}} \cdot \hat{i}$
$\rightarrow$ To obtain the magnetic field at the centre of the loop $x=0$
$\therefore B=\frac{\mu_0 IR ^2}{2 R ^3}=\frac{\mu_0 I }{2 R }$
$\rightarrow$ If there are $N$ turns, then
$\overrightarrow{ B }=\frac{\mu_0 NIR ^2}{2\left( R ^2+x^2\right)^{\frac{3}{2}}} \cdot \hat{i}$