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:(μ_o=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

The Correct Option is A

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

• μ₀​=4π×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 𝑟_𝑥𝑦= \(\^i+2\^j\)
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: 
\(B=2\sqrt2\times10^{-7} T\)