Question

A very long magnet of pole strength $16\text{Am}$ is placed vertically with its one pole on the table. At what distance from the pole, there will be a neutral point on the table
[Take $B_H=4\times {10}^{-5}{\text{Wb m}}^{-2}$]

• A. $0.2\text{m}$
• B. $0.4\text{m}$
• C. $0.6\text{m}$
• D. $0.8\text{m}$

Answer 1

The correct option is A $0.2\text{m}$


At the neutral point, horizontal component of earth's field $B_H$ exactly balances field due to the magnet.

$B=B_H$

$\frac{{\mu }_0}{4\pi }$ $\frac{m}{d^2}$ = $B_H$

$\Rightarrow {10}^{-7}\times \frac{16}{d^2}$ =$4\times {10}^{-5}$

$\Rightarrow d^2=4\times {10}^{-2}$

$\Rightarrow d=2\times {10}^{-1}\text{(or)}$

$d=0.2\text{m}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(A\right)$ is the correct answer.