Question

A long, thin-walled pipe of radius R carries a current I along its length. The current density is uniform over the circumference of the pipe. The magnetic field at the centre of the pipe due to quarter portion of the pipe is

• A. $\frac{{\mu }_{0}I\sqrt{2}}{4{\pi }^{2}R}$
• B. $\frac{{\mu }_{0}I}{{\pi }^{2}R}$
• C. $\frac{2{\mu }_{0}I\sqrt{2}}{{\pi }^{2}R}$
• D. None

Answer 1

The correct option is A $\frac{{\mu }_{0}I\sqrt{2}}{4{\pi }^{2}R}$


$dI=\frac{I}{2\pi R}\times Rd\theta =\frac{Id\theta }{2\pi }$
Now, $dB=\frac{{\mu }_{0}dI}{2\pi R}=\frac{{\mu }_{0}Id\theta }{4{\pi }^{2}R}$
$\overrightarrow{B}=\int d\overrightarrow{B}=\int dBcos\theta \hat{i}+\int dBsin\theta \hat{j}=\frac{{\mu }_{0}I}{4{\pi }^{2}R}(\int cos\theta d\theta \hat{i}+\int sin\theta d\theta \hat{j})$
Integrating anf putting limits from $0$ to $\frac{\pi }{2}$, we get
$\overrightarrow{B}=\frac{{\mu }_{0}I}{4{\pi }^{2}R}(\hat{i}+\hat{j})$
Hence, $B=\frac{{\mu }_{0}I\sqrt{2}}{4{\pi }^{2}R}$