Question

A long straight wire carrying current $16A$ is bent at $90^{\circ }$ such that half of the wire lies along the positive $x$ -axis and other half lies along the positive $y$ -axis. What is the magnitude of the magnetic field at the point, $r=(-2\hat{i}+0\hat{j})mm?($ Assume,$\frac{{\mu }_0}{4\pi }=10^{-7}H{m}^{-1})$

• A. 1.2 mT
• B. 0.8 mT
• C. 3.2 mT
• D. 1.6 mT

Answer 1

Answer: (B)

According to the question, a long straight wire is bent at $90^{\circ }$ as shown in the figure,

Where, current flowing through the wire, $I=16A$
$r=OP=(-2\hat{i}+0\hat{j})mm$
Magnitude of the magnetic field at point $P$ due to current carrying wire along $x$ -axis is zero because pont $P$ lies in the direction of conducting wire along $x$ -axis.
$\therefore$ Magnitude of the magnetic field due to current carrying wire along $y$ -axis at point $P$ is given by
$B=\frac{{\mu }_0}{4\pi }\cdot \frac{I}{r}$
Putting the given values in above relation, we get
$=\frac{{\mu }_0}{4\pi }\times \frac{16}{2\times 10^{-3}}\left[\because r=2mm=2\times 10^{-3}m\right]$
$=10^{-7}\times 8\times 10^3=0.8\times 10^{-3}T=0.8mT$