Question

A long wire carrying a current of 5 A lies along the positive Z-axis. The magnetic field the point with position vector $\vec{r}$ =($\hat{i}$+2$\hat{j}$+2$\hat{k}$) m will be:(μo=4$\pi$×10-7 in SI units)

• A. 2$\sqrt{5}$x10-7 T
• B. 5x10-7 T
• C. 0.33x10-7 T
• D. 0.66x10-7 T
• E. 7$\sqrt{5}$ x10-7 T

Answer 1

Answer: (A)

2$\sqrt{5}$x10-7 T

Answer 2

The correct answer is (A): 2$\sqrt{5}$x10-7 T
$B=\frac{μ_0​I}{2πr}$​
where:

• μ 0​=4 π ×10−7H/m (permeability of free space)
• I =5A (current through the wire)
• r is the perpendicular distance from the wire to the point where the field is being calculated

Given the position vector$r=\^{r}+2\^{j}+2\^{k}$ m, we need to find the perpendicular distance from this point to the wire, which lies along the positive z -axis.
Since the wire is along the z -axis, the distance is the magnitude of the projection of r onto the xy -plane, which is given by the vector 𝑟𝑥𝑦= $\^i+2\^j$
The magnitude of r xy ​ is:
$r=\sqrt{(1)^2+(2)^2=\sqrt{1+4}}=\sqrt5m$
Now we substitute the values into the equation for the magnetic field:
$B=\frac{μ_0​I}{2πr}$ = $\frac{(4\pi\times10^{-7})\times5}{2\pi\sqrt5}$=$=\frac{20\pi\times10^{-7}}{2\pi\sqrt5}$=$=\frac{20\times10^{-7}}{2\sqrt5}$ $=\frac{10\times10^{-7}}{\sqrt5}$
=$2 \times10^{-7}\sqrt5$
Thus, the magnetic field at the point $r=\^i+2\^j+2\^k$ is:
$B=2\sqrt2\times10^{-7} T$