A long straight wire carrying current 16 A is bent at `90^(@)` 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? (Assime, `mu_(0)/(4pi)=10^(-7) H m^(-1)`)

Options

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

Answer

B

Explanation

According to the question, a long straight wire is bent at `90^(@)` 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 point P lies in the direction of conducting wire along x-axis.
`:.` Magnitude of the magnetic field due to current carrying wire along y-axis at point P is given by
`B=mu_(0)/(4 pi).I/r`
Putting the given values in above relation, we get
`=mu_(0)/(4pi)xx16/(2xx10^(-3))" "[ :' r=2 mm =2xx10^(-3)m]`
`=10^(-7)xx8xx10^(3)=0.8xx10^(-3)=0.8xx10^(-3)T=0.8` mT