Question

A very long magnet of pole strength $4Am$ is placed vertically with its one pole on the table. The distance from the pole, the neutral point will be formed is :

$(B_H=4\times {10}^{-5}T)$

• A. $0.5m$
• B. $0.1m$
• C. $0.15m$
• D. $0.2m$

Answer 1

The correct option is B $0.1m$

At neutral point $N$ magnetic field due to $S$ pole is equal and opposite to the magnetic field due to earth.

Hence $B_S=B_H=\frac{{\mu }_{o}P}{4\pi x^2}$

where $x$ is the distance from the pole where neutral point form and $P$ is the pole strength.

putting the given values in above equations we get,

$B_S=B_H=4\times {10}^{-5}=\frac{4\pi \times {10}^{-7}\times 4}{4\pi x^2}$

solving above equation we get,

$x=0.1m$ is the distance from the pole where neutral point form.