Question

Two short magnets have equal pole strengths but one is twice as long as the other. The shorter magnet is placed 20 cm in tan A position from the compass needle. The longer magnet must be placed on the other side of the magnetometer for no deflection at a distance equal to

• A. $20\,cm$
• B. $20 \times (2)^{1/3}\, cm$
• C. $20 \times (2)^{2/3} \,cm$
• D. $20 \times (2)^{4/3} \,cm$

Answer 1

Answer: (B)

$20 \times (2)^{1/3}\, cm$

Answer 2

In tangent $A$ setting arms of the magnetometer are along east-west ( $\perp$ to magnetic meridian). Magnet is placed with its length paralleI to the arms.

As, $F = H \,tan\, \theta$
$\therefore \frac{\mu_0}{\pi} \times \frac{2M}{d^3} = H\, tan\, \theta$
For no deflection in tan A position
$\frac{\mu_0}{4 \pi} \frac{2M_1}{d_1^3} = \frac{\mu_0}{4 \pi} \frac{2M_2}{d_2^3}$
$\therefore \frac{M_1}{M_2} = \bigg(\frac{d_1}{d_2} \bigg)^3$ or $\frac{1}{2} = \bigg(\frac{20}{d_2}\bigg)^3$
or $d_2 = 20 \times (2)^{1/3}\, cm$