Question

A long straight wire (radius $=3.0mm$) carries a constant current distributed unifromly over a cross section perpandiauclar to the axis of the wire. If the current density is $100A/m^2$. The magnitude of the magnetic field at
(a) $2.0mm$ from the axis of the wire and
(b) $4.0mm$ from the axis of the wire is:

• A. $4\pi \times {10}^{-8}T,\frac{9\pi }{2}\times {10}^{-8}T$
• B. $4\pi \times {10}^{-8}T,\frac{\pi }{2}\times {10}^{-8}T$
• C. $2$$2\pi \times {10}^{-8}T,\frac{9\pi }{2}\times {10}^{-8}T$
• D. $20\pi \times {10}^{-8}T,\frac{9\pi }{2}\times {10}^{-8}T$

Answer 1

The correct option is A $4\pi \times {10}^{-8}T,\frac{9\pi }{2}\times {10}^{-8}T$

Ampere circuit law

$B\times 2\pi r={\lambda }_{w}I$

$a=3.0mm=3\times {10}^{-3}$

I= total current enclosed

in the area of circle of radius r

$I=\frac{100mp}{m^2}$

(a) area circule of radius $r=2.0mm=2\times {10}^{-3}m$ will be $A=A{R}^2=4\pi \times {10}^{-6}{m}^2$

$\Rightarrow$ current $I=IA=100\times 4\pi \times {10}^{-6}amp$

so field wire will be given by

$B\times 2\pi r={\lambda }_{w}I={\lambda }_0\times 4\pi \times {10}^{-6}\times 100$

$\Rightarrow B=\frac{2{\lambda }_w\times {10}^{-4}}{r}$=$\frac{2\times 4\pi \times {10}^{-11}}{2\times {10}^{-3}}$Tesla

putting $({\lambda }_w=4\pi \times {10}^{-7})$ m.k.s unit

or $B=4\pi \times {10}^{-8}Tesla$

(b) area circle of radius$r=4.0mm$

which is more than radius of wire $(a=3mm)$

so current =total current

$=3\times \pi (3\times {10}^{-3}{)}^2$=$9\pi \times {10}^{-4}$

field $B=\frac{{\lambda }_{0}I}{2\pi r}=\frac{4\pi \times {10}^{-7}\times 9\pi \times {10}^{-4}}{2\pi \times 4\times {10}^{-3}}$

$\frac{9\pi }{2}$$\times {10}^{-8}$ Tesla