Question

A wire carrying current $i$ is in the shape of a plane curve given by $r=f\left(\theta \right)$ where the $r$ and $\theta$ are polar coordinates. The magnetic field at the center of curvature of the current carrying wire is

• A. $\frac{{\mu }_{0}i}{4\pi }\int \frac{sin\theta d\theta }{r}$
• B. $\frac{{\mu }_{0}i}{4\pi }\int \frac{sin\theta d\theta }{r^2}$
• C. $\frac{{\mu }_{0}i}{2\pi }\int \frac{sin\theta d\theta }{r}$
• D. $\frac{{\mu }_{0}i}{2\pi }\int \frac{sin\theta d\theta }{r^2}$

Answer 1

The correct option is A $\frac{{\mu }_{0}i}{4\pi }\int \frac{sin\theta d\theta }{r}$

Here, $\text{C}$ is the center of curvature of the portion of wire.

Using biot. Savart's law:
$\overrightarrow{dB}=\frac{{\mu }_{0}i}{4\pi }\frac{\overrightarrow{dl}\times \overrightarrow{r}}{r^3}$

$\Rightarrow \text{dB}=\frac{{\mu }_{0}i(dl\times r\times sin\theta )}{4\pi r^3}$

or, $\text{dB}=\frac{{\mu }_{0}idlsin\theta }{4\pi r^2}...\left(i\right)$

From the figure shown,

$dl=rd\theta ....\left(ii\right)$

From eq. (i) and (ii)

$\text{dB}=\frac{{\mu }_{0}ird\theta sin\theta }{4\pi r^2}=\frac{{\mu }_{0}i}{4\pi r}sin\theta d\theta$

$\Rightarrow B=\int dB=\frac{{\mu }_{0}i}{4\pi }\int \frac{\sin \theta d\theta }{r}$