Question

A bar magnet of length $10cm$ and pole strength $20A-m$ is deflected through ${30}^o$ from the magnetic meridian. If earths horizontal component field is $\frac{320}{4\pi }\frac{A}{m}$. The moment of the deflecting couple is:

• A. $160\times {10}^{-7}Nm$
• B. $320\times {10}^{7}Nm$
• C. $320\times {10}^{-7}Nm$
• D. $160\pi \times {10}^{-7}Nm$

Answer 1

The correct option is A $160\times {10}^{-7}Nm$

It is given that length of bar magnet $L=10cm=0.1m$

Pole strength $m=20A-m$

earths horizontal component field $B_H=\frac{320}{4\pi }A{m}^{-1}$

$torque\tau =MB\sin \theta$

$\tau =mlB\sin \theta$

$\tau =20\times 0.1\times \frac{320}{4\pi }\times \sin 30$

$\tau =25.4Nm$