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JUNE 2024 — Physics STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 SciencePhysicsJUNE 20243 Marks
Question
Find magnetic field on the axis of a circular current carrying loop.
✓

Answer

→ As shown in the figure, a steady current I is flowing through a conducting loop of radius R .
Image

→ The loop is placed in such a way that it lies in the $y-z$ plane and the $X$-axis passing through its axis.
→ 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 .
→ 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} \ldots(1)
$
→ 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.
$
\begin{array}{l}
\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} \ldots(2)
\end{array}
$
→ 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)}
...(3)$
→ The magnetic field has two components at point P
(i) Perpendicular component $\left(d B_{\perp}=d B \sin \theta\right)$
→ When the perpendicular components are summed to get the net magnetic field, they cancel each other and the result is zero(ii) Parallel component $\left(d B_{\|}=d B \cos \theta\right)$
→ 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.
→ $
d B(x)=d B \cos \theta
$
$\therefore d B(x)=\frac{\mu_0}{4 \pi} \cdot \frac{ l d l}{ R ^2+x^2} \cdot \cos \theta \ldots$ (4) $(\because$ from equation (3))
→ From the figure,
$
\begin{array}{l}
\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}}}
\end{array}
$
→ The resultant magnetic field.
$
\begin{array}{l}
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)\left(\because \oint_{d l=2 \pi R)}\right. \\
\therefore B=\frac{\mu_0 IR^2}{2\left(R^2+x^2\right)^{\frac{3}{2}}}
\end{array}
$
→ In vector form,
$
\overrightarrow{B}=\frac{\mu_0 IR^2}{2\left(R^2+x^2\right)^{\frac{3}{2}}} \cdot \hat{i}
$
→ To obtain the magnetic field at the centre of the loop $x=0$
$
\therefore B=\frac{\frac{\mu_0 IR^2}{2 R^3}=\frac{\mu_0 I}{2 R}, ~}{2 R}
$
→ If there are N turns, then
$
\overrightarrow{B}=\frac{\mu_0 NIR^2}{2\left(R^2+x^2\right)^{\frac{3}{2}}}
$

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