Question

A long straight wire carrying current $16\, A$ 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, $r =(-2 \hat{ i }+0 \hat{ j }) mm ?\left(\right.$ Assume,$\left.\frac{\mu_{0}}{4 \pi}=10^{-7} H\, m ^{-1}\right)$

• 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, $I=16\, A$
$r = OP =(-2 \hat{ i }+0 \hat{ j }) mm$
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=2 \,mm =2 \times 10^{-3} m \right] $
$=10^{-7} \times 8 \times 10^{3}=0.8 \times 10^{-3} T =0.8\, mT$