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Consider a circular current-carrying loop of radius $\mathrm{R}$ in the $\mathrm{x}-\mathrm{y}$ plane with centre at origin. Consider the line integral
$$
\mathfrak{J}(\mathrm{L})=\left|\int_{-\mathrm{L}}^{\mathrm{L}} \mathrm{B} d l\right|
$$
taken along z-axis.
(a) Show that $\mathfrak{J}(\mathrm{L})$ monotonically increases with $\mathrm{L}$
(b) Use an appropriate amperian loop to show that $\mathfrak{J}(\infty)=\mu_0 I$. where I is the current in the wire
(c) Verify directly the above result
(d) Suppose we replace the circular coil by a square coil of sides $R$ carrying the same current $I$.
What can you say about $\mathfrak{J}(\mathrm{L})$ and $\mathfrak{J}(\infty)$ ?
$$
\mathfrak{J}(\mathrm{L})=\left|\int_{-\mathrm{L}}^{\mathrm{L}} \mathrm{B} d l\right|
$$
taken along z-axis.
(a) Show that $\mathfrak{J}(\mathrm{L})$ monotonically increases with $\mathrm{L}$
(b) Use an appropriate amperian loop to show that $\mathfrak{J}(\infty)=\mu_0 I$. where I is the current in the wire
(c) Verify directly the above result
(d) Suppose we replace the circular coil by a square coil of sides $R$ carrying the same current $I$.
What can you say about $\mathfrak{J}(\mathrm{L})$ and $\mathfrak{J}(\infty)$ ?
Solution:
2254 Upvotes
Verified Answer
By using Ampere circuital law.
(a) B (z) points in the same direction on $\mathrm{z}$-axis and hence, $\mathrm{J}(\mathrm{L})$ is a monotonical function of L. As B and dl are along the same direction, so $\vec{B} . \mathrm{d} \vec{l}=B$.dl $\cos \theta=B$. $d l$ $\cos 0^{\circ}=\mathrm{B} . \mathrm{d} 1$
(b) $\mathrm{J}(\mathrm{L})+$ contribution from large distance on contour $\mathrm{C}=\mu_0 \mathrm{I}$
As, $\quad \mathrm{L} \rightarrow \infty$
Contribution from large distance $\rightarrow 0$ as
$$
\begin{aligned}
&\mathrm{B} \propto 1 / \mathrm{r}^3 \\
&\mathrm{~J}(\infty)-\mu_0 \mathrm{I}
\end{aligned}
$$
(c) The magnetic field due to circular current-carrying loop of radius R in the $x-y$ plane with centre at origin at any point lying at a distance a from origin.
Let, a point $P$ is in $\mathrm{z}$-axis at distacnez, i.e, $\mathrm{OP}=\mathrm{Z}$ and angle between $R$ and $Q P=\theta$ then the magnetic field at $P$ due to loop is
$$
\begin{aligned}
&\mathrm{B}_{\mathrm{z}}=\frac{\mu_0 1 \mathrm{R}^2}{2\left(\mathrm{z}^2+\mathrm{R}^2\right)^{3 / 2}} \\
&\int_{-\infty}^{\infty} B_z d z=\int_{-\infty}^{\infty} \frac{\mu_0 1 R^2}{2\left(z^2+R^2\right)^{3 / 2}} d z \quad\left(\because \tan \theta=\frac{z}{R}\right) \\
&\begin{array}{lc}
\text { So, } & \mathrm{z}=\mathrm{R} \tan \theta \\
\Rightarrow & \mathrm{dz}=\mathrm{R} \sec ^2 \theta \mathrm{d} \theta
\end{array} \\
&
\end{aligned}
$$
(d) $\mathrm{B}(\mathrm{z})_{\text {square loop }} < $ loop B(z) circular loop
$\therefore \quad(\mathrm{L})_{\text {square loop }},=(\mathrm{L})_{\text {circular loop }}$
By using arguments as in (b)
$(\infty)_{\text {square loop }}=(\infty)_{\text {circular loop }}$
(a) B (z) points in the same direction on $\mathrm{z}$-axis and hence, $\mathrm{J}(\mathrm{L})$ is a monotonical function of L. As B and dl are along the same direction, so $\vec{B} . \mathrm{d} \vec{l}=B$.dl $\cos \theta=B$. $d l$ $\cos 0^{\circ}=\mathrm{B} . \mathrm{d} 1$
(b) $\mathrm{J}(\mathrm{L})+$ contribution from large distance on contour $\mathrm{C}=\mu_0 \mathrm{I}$
As, $\quad \mathrm{L} \rightarrow \infty$
Contribution from large distance $\rightarrow 0$ as
$$
\begin{aligned}
&\mathrm{B} \propto 1 / \mathrm{r}^3 \\
&\mathrm{~J}(\infty)-\mu_0 \mathrm{I}
\end{aligned}
$$
(c) The magnetic field due to circular current-carrying loop of radius R in the $x-y$ plane with centre at origin at any point lying at a distance a from origin.
Let, a point $P$ is in $\mathrm{z}$-axis at distacnez, i.e, $\mathrm{OP}=\mathrm{Z}$ and angle between $R$ and $Q P=\theta$ then the magnetic field at $P$ due to loop is
$$
\begin{aligned}
&\mathrm{B}_{\mathrm{z}}=\frac{\mu_0 1 \mathrm{R}^2}{2\left(\mathrm{z}^2+\mathrm{R}^2\right)^{3 / 2}} \\
&\int_{-\infty}^{\infty} B_z d z=\int_{-\infty}^{\infty} \frac{\mu_0 1 R^2}{2\left(z^2+R^2\right)^{3 / 2}} d z \quad\left(\because \tan \theta=\frac{z}{R}\right) \\
&\begin{array}{lc}
\text { So, } & \mathrm{z}=\mathrm{R} \tan \theta \\
\Rightarrow & \mathrm{dz}=\mathrm{R} \sec ^2 \theta \mathrm{d} \theta
\end{array} \\
&
\end{aligned}
$$
(d) $\mathrm{B}(\mathrm{z})_{\text {square loop }} < $ loop B(z) circular loop
$\therefore \quad(\mathrm{L})_{\text {square loop }},=(\mathrm{L})_{\text {circular loop }}$
By using arguments as in (b)
$(\infty)_{\text {square loop }}=(\infty)_{\text {circular loop }}$
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