Question

A short magnet of moment $6.75\, Am ^{2}$ produces a neutral point on its axis. If horizontal component of earth's magnetic field is $5 \times 10^{-5}\, Wb / m ^{2}$, then the distance of the neutral point should be

• A. 10 cm
• B. 20 cm
• C. 30 cm
• D. 40 cm

Answer 1

Answer: (C)

At neutral point,
$\left| \text{Magnetic field due to magnet}\right|= \left| \text{Magnetic field due to earth}\right|$
$\frac{\mu_{0}}{4 \pi} \cdot \frac{2 M}{d^{3}}=5 \times 10^{-5}$
$\Rightarrow 10^{-7} \times \frac{2 \times 6.75}{d^{3}}=5 \times 10^{-5}$
$d=0.3\, m =30\, cm$