Question

A long magnet is placed vertically with its $S$ -pole resting on the table. A neutral point is obtained $10\, cm$ from the pole due geographic north of it. If $H=3.2 \times 10^{-5} T$, then the pole strength of magnet is

• A. $8\, ab - A-cm ^{-1}$
• B. $16\, ab - A- cm ^{-1}$
• C. $32\, ab - A - cm ^{-1}$
• D. $64\, ab - A - cm ^{-1}$

Answer 1

Answer: (C)

As, the magnet is long, we assume that the upper north pole produces no effect. But due to south pole of the magnet is equal and opposite to horizontal component of earth's magnetic field, i.e.,
$B=\left(\frac{\mu_{0}}{4 \pi}\right) \frac{m}{r^{2}}=H$
In CGS system, $\frac{\mu_{0}}{4 \pi}=1$
$\therefore 1 \times \frac{m}{10^{2}} =3.2 \times 10^{-5} \times 10^{4}$ (gauss)
$m =32\, ab - amp - cm ^{-1}$