Question

A dipole of magnetic moment $2A-m^2$ is deflected through ${30}^o$ from magnetic meridian. If magnetic field strength of Earth's horizontal component is $H=30A/m$, the deflecting couple is

• A. $3.8\times {10}^{-5}N-m$
• B. $38\times {10}^{-5}N-m$
• C. $1.9\times {10}^{-5}N-m$
• D. $38\times {10}^{-7}N-m$

Answer 1

The correct option is A $3.8\times {10}^{-5}N-m$

$m=2A{m}^2$
$\theta ={30}^o$
$H=30A/m$
$B_H={\mu }_{o}H$
$={\mu }_{o}30$
$z=mB\sin {30}^o$
$=2\times 4\pi \times {10}^{-7}\times 30\times \frac{1}{2}$
$=3.769\times {10}^{-5}Nm$