Question

A long straight wire (radius $=3.0mm)$ carries a constant current distributed uniformly over a cross section perpendicular to the axis of the wire. If the current density is 100$A/m^2$ . The magnitudes of the magnetic field at (a) 2.0$mm$ from the axis of the wire and 4.0$mm$ from the axis of the wire is

• A. $1.4e-7T$
• B. $4\pi \times {10}^{-8}T,\frac{\pi }{2}\times {10}^{-8}T$
• C. $4\pi \times {10}^{-8}T,\frac{9\pi }{2}\times {10}^{-8}T$
• D. $\pi {10}^{-4}T,\frac{9\pi }{2}\times {10}^{-8}T$

Answer 1

The correct option is A $1.4e-7T$

$\begin{array}{c}\left(a\right)I=JA\\ I=J\left(\pi r^2\right)\\ \int B.dL=\left({\mu }_0\right)I\\ B\left(2\pi r\right)=\left({\mu }_0\right)\left(J\right)\left(\pi r^2\right)\\ B=\frac{\left({\mu }_0\right)\left(J\right)}{2}\\ B=\frac{\left(\frac{1.257e-6}{A^2}\right)\left(\frac{100A}{m^2}\right)\left(\frac{2e-3m}{2}\right)}{2}\\ B=1.3e-7T\\ \left(b\right)I=JA\\ I=J\left(\pi R^2\right)\\ \int B.dL=\left({\mu }_0\right)I\\ B\left(2\pi r\right)=\left({\mu }_0\right)\left(J\right)\left(\pi R^2\right)\\ B=\frac{\left({\mu }_0\right)\left(J\right)\left(R^2\right)}{2r}\\ B=\frac{\left(\frac{1.257e-6N}{A^2}\right)\left(\frac{100A}{m^2}\right){\left(3e-3m\right)}^2}{\left(2\times 4e-3m\right)}\\ B=1.4e-7T\end{array}$

Hence, Option $A$ is correct.