Question

The magnetic moment of a small current carrying loop is $2.1\times {10}^{-25}amp\times m^2.$ The magnetic field at a point on its axis at a distance of $1\stackrel{\circ }{A}$ is

• A. $4.2\times {10}^{-2}weber/m^2$
• B. $4.2\times {10}^{-3}weber/m^2$
• C. $4.2\times {10}^{-4}weber/m^2$
• D. $4.2\times {10}^{-5}weber/m^2$

Answer 1

The correct option is A $4.2\times {10}^{-2}weber/m^2$

Field at point x from the centre of a current carrying loop on the axis is
$B=\frac{{\mu }_0}{4\pi }.\frac{2M}{x^3}=\frac{{10}^{-7}\times 2\times 2.1\times {10}^{-25}}{({10}^{-10}{)}^3}=4.2\times {10}^{-32}\times {10}^{30}=4.2\times {10}^{-2}W/m^2$