Question

A bar magnet of $5$ cm long having a pole strength of $20$ A.m. is deflected through ${30}^0$ from the magnetic meridian. If $H=\frac{320}{4\pi }$ A/m, the deflecting couple is

• A. $1.6\times {10}^{-4}$ Nm
• B. $3.2\times {10}^{-5}$ Nm
• C. $1.6\times {10}^{-5}$ Nm
• D. $1.6\times {10}^{-2}$ Nm

Answer 1

Answer: (C)

$1.6\times {10}^{-5}$ Nm

$l=5m$
$P=20Am$
$N=\frac{320}{4\pi }A/m$
$B=N{\mu }_0$
$=320\times {10}^{-7}T$
$\overrightarrow{z}=\overrightarrow{m}\times \overrightarrow{B}\theta ={30}^0$
$z=mBsin\theta$
$z=\frac{5}{100}\times 20\times 320\times {10}^{-7}\times sin30$
$z=1.6\times {10}^{5}Nm$