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Magnetic Fields due to Electric Current — Physics STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 SciencePhysicsMagnetic Fields due to Electric Current4 Marks
Question
Obtain an expression for the magnetic induction produced by a current in a wire in the shape of a circular arc at its centre of curvature. Hence obtain an expression for the magnetic induction at the centre of a circular coil carrying a current.
✓

Answer

Consider a wire in the shape of a circular arc of radius of curvature $R$ and carrying a current I. The unit vector $\hat{ r }$ from each current element I $\overrightarrow{d l}$ towards
Image

the centre of the loop is perpendicular to $\overrightarrow{I d l}$, i.e., the angle $\theta$ between them is $90^{\circ}$. The direction of the incremental magnetic induction $\overrightarrow{d B}$ due to each current element is in the same direction, viz., perpendicular to the plane of the loop, and out of the plane of the figure for the sense of the current shown in above figure.
Since every current element is equidistant from the centre of curvature $C$, the magnetic field at $C$ due to each current element in the arc by Biot-Savart's law is
$
d \vec{B}=\left(\frac{\mu_0}{4 \pi}\right) \frac{I \overrightarrow{d l} \times \hat{ r }}{r^2}....(1)
$
In magnitude,
$
\begin{aligned}
d B & =\left(\frac{\mu_0}{4 \pi}\right) \frac{I d l \sin \theta}{r^2} \\
& =\left(\frac{\mu_0}{4 \pi}\right) \frac{I d l \sin 90^{\circ}}{r^2}=\left(\frac{\mu_0}{4 \pi}\right) \frac{I d l}{r^2}...(2)
\end{aligned}
$
If the arc subtends an angle 0 at its centre of curvature $C$, the total field at $C$ due to all the elements on the arc is
$
\begin{aligned}
B & =\int d B=\int_0^\phi \frac{\mu_0}{4 \pi} \frac{i R d \phi}{R^2}=\frac{\mu_0 i}{4 \pi R}=\int_0^\phi d \phi \\...(3)
& =\frac{\mu_0}{4 \pi} \frac{I \phi}{R}....(4)
\end{aligned}
$
where $\Phi$ is in radian.
The magnitude of the total induction $\vec{B}$ at the centre of a circular coil is, from Eq. (3),
$
B=\frac{\mu_0 I}{4 \pi R} \int_0^{2 \pi} d \phi=\frac{\mu_0 I}{4 \pi R}(2 \pi)=\frac{\mu_0 I}{2 R}
$
Image
If a circular coil has $N$ turns, each of radius $r$ and carries a current $I$, the magnetic induction at its centre has a magnitude
$
B =\frac{\mu_0 N I}{2 R}..(6)
$

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