Answer:
The magnetic field is 1.6 nT in z-direction
Explanation:
Given;
Length of wire, L = 2.0-cm
Current in the wire, I = 20-A
point from the axis, r = 5.0 m
Strength of magnetic field can be calculated by applying Biot-Savart equation;
[tex]B = \frac{\mu I}{2\pi r} (\frac{L}{\sqrt{4r^2 } +L} )[/tex]
where;
μ is constant given as 4π x 10⁻⁷ T.m/A
Substitute the given values in the equation above and calculate strength of magnetic field.
Since current is moving in y-direction and magnetic field will be caculated from a point 5m in the x-direction, based on vector law, the resultant direction will be z.
[tex]B = \frac{\mu I}{2\pi r} (\frac{L}{\sqrt{4r^2 } +L} )\\\\B = \frac{4\pi *10^{-7} *20}{2\pi *5} (\frac{0.02}{\sqrt{4(5)^2 } +0.02} )\\\\B = 8*10^{-7} *1.9998 *10^{-3}\\\\B = 1.6*10^{-9} \ T\\\\B = 1.6 \ nT[/tex]
Therefore, the magnetic field is 1.6 nT in z-direction
