Answer: (a) B = 2 x 10⁻⁵T
(b) B = 1.94 x 10⁻⁵T
(c) B = 1.8 x 10⁻⁴T
Explanation: A magnetic field due to a current passing through a straight wire is calculated using the Biot-Savart Law:
[tex]dB=\frac{\mu_{0}}{4.\pi} \frac{IdLXR}{r^{3}}[/tex]
where
dL is current length element
[tex]\mu_{0}[/tex] is permeability of free space ([tex]4.\pi.10^{-7}[/tex]T.m/A)
(a) For a infinite straight wire:
[tex]B=\frac{\mu_{0}I}{2.\pi.R}[/tex]
[tex]B=\frac{4.\pi.10^{-7}.2}{2.\pi.2.10^{-2}}[/tex]
B = 2x10⁻⁵T
For an infinite, long and straight wire, magnetic field is 2x10⁻⁵T.
(b) For a finite wire:
[tex]B=\frac{\mu_{0}I}{2.\pi.R}\frac{L}{\sqrt{L^{2}+R^{2}} }[/tex]
[tex]B=\frac{4.\pi.10^{-7}.2}{2.\pi.2.10^{-2}} \frac{8.10^{-2}}{\sqrt{(8.10^{-2})^{2}+(2.10^{-2})^{2}} }[/tex]
B = 1.94x10⁻⁵T
The magnetic field for a finite wire in the same conditionsas infinite wire is 1.94x10⁻⁵T.
(c) For a finite wire at a point distant from the end of the wire:
[tex]B=\frac{\mu_{0}I}{4.\pi.L\sqrt{2} }[/tex]
[tex]B=\frac{4.\pi.10^{-7}.2}{4.\pi.8.10^{-2}\sqrt{2} }[/tex]
B = 0.18x10⁻⁵T
At a point at the end, magnetic field is 1.8x10⁻⁴T.