Question

A small bar magnet has a magnetic moment $1.2A-m^2.$ The magnetic field at a distance $0.1m$ on its axis will be : $({\mu }_0=4\pi \times {10}^{-7}Tm/A)$

• A. $1.2\times {10}^{-4}T$
• B. $2.4\times {10}^{-4}T$
• C. $2.4\times {10}^{4}T$
• D. $1.2\times {10}^{4}T$

Answer 1

Answer: (B)

$2.4\times {10}^{-4}T$

Here $M=1.2A-m^2$ , $d=0.1m$ , ${\mu }_o=4\pi \times {10}^{-7}Tm/A$

On axis we know that,

$B=\frac{{\mu }_{0}2M}{4\pi d^3}=\frac{4\pi \times {10}^{-7}\times 2\times 1.2}{4\pi (0.1{)}^3}=2.4\times {10}^{-4}T$